law of acceleration problem solving worksheet with answers

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F=ma Practice Problems

F=ma problem set.

Practice solving for net force, using Newtons second law (F=ma), and relating F=ma to the acceleration equations.

In these practice problems we will either use F=ma or our 1D motion acceleration equations to solve force problems.

1. What is the acceleration of the 15 kg box that has 500 N of force applied to the right?

a = 33.33 m/s 2 Right

2. What is the acceleration of the 25 kg box that has 50 N of force applied to the right?

                a=2.0 m/s 2 Right

3. What is the acceleration of the 3 kg box that has 25 N of force applied to the right and 55 N left?

               a = 10.0 m/s 2 Left

4. What is the acceleration of the 5 kg box that has a 25 N force and 50 N force applied both right?

            a = 15.0 m/s 2 Right

5. What is the acceleration of the 25 kg box that has a 100 N force north and 50 N force east applied?

             a= 4.47 m/s 2

5b. What direction would this box accelerate?

            63.43° North of East

6. Does a 795 kg Lorinser speedy, 6300 kg elephant, or 8.6 kg wagon have more inertia and why?

            6300 kg Elephant

            The more mass the more inertia

7. How much force is required to accelerate a 795 kg Lorinser Speedy by 15 m/s 2 ?

            F = 11925 N

8. How much force is required to accelerate an 8.6 kg wagon by 15 m/s 2 ?

            F = 129 N

9. How much does a 6300 kg elephant accelerate when you apply 500 N of force?

            a = 0.0794 m/s 2

10. What is the mass of an object if it takes a net force of 40 N to accelerate at a rate of 0.88 m/s 2 ?

            m = 45.45 kg

11. How much force is required to accelerate a 0.142 kg baseball to 44.7 m/s during a pitchers 1.5 meter delivery?

            F = 94.58 N

12. A 0.050 kg golf ball leaves the tee at a speed of 75.0 m/s. The club is in contact with the ball for 0.020 s. What is the net force of the club on the ball?

                F = 187.5 N

13. A 90.0 kg astronaut receives a 30.0 N force from her jetpack. How much faster is she be moving after 2.00 seconds?

            0.667 m/s faster

14. A 795 kg car starts from rest and travels 41 m in 3.0 s. How much force did the car engine provide?

            F = 7242 N

15. Joe and his sailboat have a combined weight of 450 kg. How far has Joe sailed when he started at 5 m/s and a gust of wind provided 600 Newtons of force for 4 seconds?

            x = 30.64 m

16. Tom pulls a 45 kilogram wagon with a force of 200 Newtons at a 15° angle to the horizontal from rest. How much faster will the wagon be moving after 2 seconds?

            v f = 8.58 m/s

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Newton Second Law of Motion Example Problems with Answers

Newton's 2nd law of motion involves force, mass and acceleration of an object. It is the acceleration of an object produced by an action or force which is directly proportional to the magnitude of the net force in the same direction and inversely proportional to the object mass. Calculate net force, mass and acceleration of an object by referring the below Newton second law of motion example problems with answers.

Newton 2nd Law of Motion Problems with Solutions

Let us consider the problem: A 15 kg object moving to the west with an acceleration of 10m/s 2 . What is the net force acting on an object?

We can calculate Force, Mass and Acceleration using the given formula.

Newton's Second Law of Motion Formulae:

Substituting the values in the above given formula,

Net Force (F net )= 15 x 10 = 150 N Therefore, the value of Net force = 150 N

Refer the newton 2nd law of motion problems with solutions: A softball has a mass of 1.5 kg and hits the catcher's glove with a force of 30 N? What is the acceleration of the softball?

Substituting the values in the above given formula, Acceleration = 30 / 1.5 = 20 m/s 2 Therefore, the value of Acceleration is 20 m/s 2

Refer the problem with solution: What is the mass of a truck if it produces a force of 15000 N while accelerating at a rate of 6 m/s 2 ?

Substituting the values in the above given formula, Mass = 15000 / 6 = 2500 kg Therefore, the value of Mass is 2500 kg

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6.1 Solving Problems with Newton’s Laws

Learning objectives.

By the end of this section, you will be able to:

• Apply problem-solving techniques to solve for quantities in more complex systems of forces
• Use concepts from kinematics to solve problems using Newton’s laws of motion
• Solve more complex equilibrium problems
• Solve more complex acceleration problems
• Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

• Identify the physical principles involved by listing the givens and the quantities to be calculated.
• Sketch the situation, using arrows to represent all forces.
• Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
• Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
• Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain

Significance

Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

• We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
• Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

• We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
• Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

• ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
• Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Interactive

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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Newton’s second law of motion – problems and solutions

Solved problems in Newton’s laws of motion – Newton’s second law of motion 

1. A 1 kg object accelerated at a constant 5 m/s 2 . Estimate the net force needed to accelerate the object.

Mass (m) = 1 kg

Acceleration (a) = 5 m/s 2

Wanted : net force (∑F)

We use Newton’s second law to get the net force.

∑ F = (1 kg)(5 m/s 2 ) = 5 kg m/s 2 = 5 Newton

2. Mass of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….

Net force (∑F) = 2 Newton

Wanted : The magnitude and direction of the acceleration (a)

a = 2 m/s 2

The direction of the acceleration = the direction of the net force (∑F)

3. Object’s mass = 2 kg, F 1 = 5 Newton, F 2 = 3 Newton. The magnitude and direction of the acceleration is…

Mass (m) = 2 kg

F 1 = 5 Newton

F 2 = 3 Newton

net force :

∑ F = F 1 – F 2 = 5 – 3 = 2 Newton

The magnitude of the acceleration :

a = 1 m/s 2

Direction of the acceleration = direction of the net force = direction of F 1

4. Object’s mass = 2 kg, F 1 = 10 Newton, F 2 = 1 Newton. The magnitude and direction of the acceleration is…

F 2 = 1 Newton

F 1 = 10 Newton

F 1x = F 1 cos 60 o = (10)(0.5) = 5 Newton

Net force :

∑ F = F 1x – F 2 = 5 – 1 = 4 Newton

Direction of the acceleration = direction of the net force = direction of F 1x

5. F 1 = 10 Newton, F 2 = 1 Newton, m 1 = 1 kg, m 2 = 2 kg. The magnitude and direction of the acceleration is…

Mass 1 (m 1 ) = 1 kg

Mass 2 (m 2 ) = 2 kg

The net force :

∑ F = F 1 – F 2 = 10 – 1 = 9 Newton

a = ∑F / (m 1 + m 2 )

a = 9 / (1 + 2)

a = 3 m/s 2

The direction of the acceleration = the direction of the net force = direction of F 1

A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/ s 2 . Determine the magnitude of friction force experienced by the block.

Mass (m) = 40 kg

Force (F) = 200 N

Acceleration (a) = 3 m/s 2

Wanted: Friction force (F g )

The equation of Newton’s second law of motion

∑ F = net force, m = mass, a = acceleration

The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).

Choose rightward as positive and leftward as negative.

F – F g = m a

200 – F g = (40)(3)

200 – F g = 120

F g = 200 – 120

F g = 80 Newton

The correct answer is D.

7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the normal force exerted by block B on block A.

Force (F) = 5 Newton

Mass of block A (m A ) = 100 gram = 0.1 kg

Mass of block B (m B ) = 300 gram = 0.3 kg

Acceleration of gravity (g) = 10 m/s 2

Weight of block A (w A ) = (0.1 kg)(10 m/s 2 ) = 1 kg m/s 2 = 1 Newton

Weight of block B (w B ) = (0.3 kg)(10 m/s 2 ) = 3 kg m/s 2 = 3 Newton

Wanted : Normal force exerted by block B to block A

F = push force (act on block B)

w A = weight of block A (act on block A)

w B = weight of block B (act on block B)

N A = normal force exerted by block B on block A (Act on block A)

N A ’ = normal force exerted by block A on block B (Act on block B)

Apply Newton’s second law of motion on both blocks :

F – w A – w B + N A – N A ’ = (m A + m B ) a

N A and N A ’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.

F – w A – w B = (m A + m B ) a

5 – 1 – 3 = (0.1 + 0.3) a

5 – 4 = (0.4) a

1 = (0.4) a

a = 1 / 0.4

a = 2.5 m/s 2

Apply Newton’s second law of motion on block A :

N A – w A = m A a

N A – 1 = (0.1)(2.5)

N A – 1 = 0.25

N A = 1 + 0.25

N A = 1.25 Newton

The correct answer is B.

8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.

B. 4 N downward

C. 9 N upward

D. 9 N downward

Weight of X (w X ) = 4 Newton

Pull force (F x ) = 2 Newton

Tension force (F T ) = 9 Newton

Wanted: Net force acts on object X

Vertically upward forces that act on object X :

The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.

Vertically downward forces that act on object X :

There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight w x and the horizontal component of force F x .

Net force act on the object X :

F T – w X – F x = 9 – 4 – 2 = 9 – 6 = 3

The net force act on the object X is 3 Newton, vertically upward.

The correct answer is A.

9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s 2 . If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.

A. 1.75 m/s 2

B. 1.50 m/s 2

C. 1.00 m/s 2

D. 0.88 m/s 2

Force (F) = 16 Newton = 16 kg m/s 2

Acceleration (a) = 2 m/s 2

Friction force (F fric ) = 2 Newton = 2 kg m/s 2

Wanted : Object’s acceleration ?

Smooth horizontal surface (no friction force) :

Mass of object is 8 kilogram.

Rough horizontal surface (there is a friction force) :

F – F fric = m a

16 – 2 = 8 a

a = 1.75 m/s 2

Object’s acceleration is 1.75 m/s 2 .

10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms -2 , then determine the magnitude and direction of force act by Tom.

A. 1.70 N and its direction is opposite with force acted by Andre.w

B. 1.70 N and its direction same as force acted by Andrew

C. 2.30 N and its direction is opposite with force acted by Andrew.

D. 2.30 N and its direction same as force acted by Andrew.

Push force acted by Andrew (F 1 ) = 5.70 Newton

Mass of object (m) = 2.00 kg

Acceleration (a) = 2.00 m/s 2

Wanted : Magnitude and direction of force acted by Tom (F 2 ) ?

Apply Newton’s second law of motion :

F 1 + F 2 = m a

5.70 + F 2 = (2)(2)

5.70 + F 2 = 4

F 2 = 4 – 5.70

F 2 = – 1.7 Newton

Minus sign indicated that (F 2 ) is opposite with push force act by Andrew (F 1 ).

11. If the mass of the block is the same, which figure shows the smallest acceleration?

Net force A :

ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward

Net force B :

ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward

Net force C :

ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward

Net force D :

ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward

The equation of Newton’s second law :

a = acceleration, ΣF = net force, m = mass

Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration. Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.

12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.

Determine the object’s acceleration.

Mass of object (m) = 20 kg

Net force (ΣF) = 25 N + 30 N – 15 N = 40 N

Wanted: Acceleration of an object

Object’s acceleration calculated using the equation of Newton’s second law :

a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s 2

13. Which statements below describes Newton’s third law?

(1) Passengers pushed forward when the bus braked suddenly

(2) B ooks on paper are not falling when the paper is pulled quickly

(3) When playing skateboard when the foot pushes the ground back then the skateboard will slide forward

(4) O ars pushed backward, boats moving forward

(1) Newton’s first law

(2) Newton’s first law

(3) Newton’s third law

(4) Newton’s third law

[wpdm_package id=’470′]

• Mass and weight
• Normal force
• Newton’s second law of motion
• Friction force
• Motion on the horizontal surface without friction force
• The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force
• Motion on the inclined plane without friction force
• Motion on the rough inclined plane with the friction force
• Motion in an elevator
• The motion of bodies connected by cord and pulley
• Two bodies with the same magnitude of accelerations
• Rounding a flat curve – dynamics of circular motion
• Rounding a banked curve – dynamics of circular motion
• Uniform motion in a horizontal circle
• Centripetal force in uniform circular motion

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Acceleration

Practice problem 1.

• A car is said to go "zero to sixty in six point six seconds". What is its acceleration in m/s 2 ?
• The driver can't release his foot from the gas pedal (a.k.a. the accelerator). How many additional seconds would it take for the driver to reach 80 mph assuming the aceleration remains constant?
• A car moving at 80 mph has a speed of 35.8 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?

Well first of all, we shouldn't be dealing with English units. They're difficult to work with, so let's convert them straight away and then do the old "plug and chug".

Since the question asked for acceleration and acceleration is a vector quantity this answer is not complete. A proper answer must include a direction as well. This is quite easy to do. Since the car is starting from rest and moving forward, its acceleration must also be forward. The ultimate, complete answer to this problem is the car is accelerating at…

a  =  4.06 m/s 2  forward

We should convert the final speed to SI units.

Use the fact that change equals rate times time, and then add that change to our velocity at the end of the previous problem. Algebra will do the rest for us.

Alternate solution. We don't need no stinkin' conversions with this method. The ratio of eighty to sixty is a simple one, namely 4 3 . From our definition of acceleration, it should be apparent that time is directly proportional to change in velocity when acceleration is constant. Thus…

This is not the answer. It is the time elapsed from the moment when the car began to move. The question was about the additional time needed, so we should subtract the time required to go from zero to sixty. Thus…

∆ t  =  8.8 s − 6.6 s  =  2.2 s

The two methods give essentially the same answer.

Quite simple. Let's do it.

Nothing surprising there except the negative sign. When a vector quantity is negative what does it mean? There are several interpretations of this, but I think mine is the best. When a vector has a negative value, it means that it points in a direction opposite that of the positive vectors. In this problem, since the positive vectors are assumed to point forward (What other direction would a normal car drive?) the acceleration must be backward. Thus the complete answer to this problem is that the car's acceleration is…

a  =  7.16 m/s 2 backward

Although it is common to assign deceleration a negative value, negative acceleration does not automatically imply deceleration. When dealing with vector quantities, any direction can be assumed positive…

up, down, right, left, forward, backward, north, south, east, west

and the corresponding opposite direction assumed negative…

down, up, left, right, backward, forward, south, north, west, east.

It won't matter which you chose as long as you are consistent throughout a problem. Don't learn any rules for assigning signs to particular directions and don't let anyone tell you that a certain direction must be positive or must be negative.

practice problem 2

Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign. Every quantity that points away from the batter will be positive. Every quantity that points toward him will be negative. Thus, the ball comes in at −40 m/s and goes out at +50 m/s. If we didn't pay attention to this detail, we wouldn't get the right answer.

practice problem 3

Practice problem 4.

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6.3: Solving Problems with Newton's Laws (Part 2)

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Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6: What Force Must a Soccer Player Exert to Reach Top Speed?

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What is her average acceleration? (b) What average force does the ground exert forward on the runner so that she achieves this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible.

To find the answers to this problem, we use the problem-solving strategy given earlier in this chapter. The solutions to each part of the example illustrate how to apply specific problem-solving steps. In this case, we do not need to use all of the steps. We simply identify the physical principles, and thus the knowns and unknowns; apply Newton’s second law; and check to see whether the answer is reasonable.

• We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is \(\Delta\)v = 8.00 m/s . We are given the elapsed time, so \(\Delta\)t = 2.50 s. The unknown is acceleration, which can be found from its definition: $$a = \frac{\Delta v}{\Delta t} \ldotp$$Substituting the known values yields $$a = \frac{8.00\; m/s}{2.50\; s} = 3.20\; m/s^{2} \ldotp$$
• Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, $$F_{net} = ma \ldotp$$Substituting the known values of m and a gives $$F_{net} = (70.0\; kg)(3.20\; m/s^{2}) = 224\; N \ldotp$$

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Significance

This example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles, the knowns, and the unknowns involved in the problem. The second step is to solve for the unknown, in this case using Newton’s second law. Finally, we check our answer to ensure it is reasonable. These techniques for integrated concept problems will be useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life.

Exercise 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7: What Force Acts on a Model Helicopter?

A 1.50-kg model helicopter has a velocity of 5.00 \(\hat{j}\) m/s at t = 0. It is accelerated at a constant rate for two seconds (2.00 s) after which it has a velocity of (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\)) m/s. What is the magnitude of the resultant force acting on the helicopter during this time interval?

We can easily set up a coordinate system in which the x-axis (\(\hat{i}\) direction) is horizontal, and the y-axis (\(\hat{j}\) direction) is vertical. We know that \(\Delta\)t = 2.00s and \(\Delta\)v = (6.00 \(\hat{i}\) + 12.00 \(\hat{j}\) m/s) − (5.00 \(\hat{j}\) m/s). From this, we can calculate the acceleration by the definition; we can then apply Newton’s second law.

\[a = \frac{\Delta v}{\Delta t} = \frac{(6.00 \hat{i} + 12.00 \hat{j}\; m/s) - (5.00 \hat{j}\; m/s)}{2.00\; s} = 3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}$$ $$\sum \vec{F} = m \vec{a} = (1.50\; kg)(3.00 \hat{i} + 3.50 \hat{j}\; m/s^{2}) = 4.50 \hat{i} + 5.25 \hat{j}\; N \ldotp\]

The magnitude of the force is now easily found:

\[F = \sqrt{(4.50\; N)^{2} + (5.25\; N)^{2}} = 6.91\; N \ldotp\]

The original problem was stated in terms of \(\hat{i}\) − \(\hat{j}\) vector components, so we used vector methods. Compare this example with the previous example.

Exercise 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8: Baggage Tractor

Figure \(\PageIndex{7}\)(a) shows a baggage tractor pulling luggage carts from an airplane. The tractor has mass 650.0 kg, while cart A has mass 250.0 kg and cart B has mass 150.0 kg. The driving force acting for a brief period of time accelerates the system from rest and acts for 3.00 s. (a) If this driving force is given by F = (820.0t) N, find the speed after 3.00 seconds. (b) What is the horizontal force acting on the connecting cable between the tractor and cart A at this instant?

A free-body diagram shows the driving force of the tractor, which gives the system its acceleration. We only need to consider motion in the horizontal direction. The vertical forces balance each other and it is not necessary to consider them. For part b, we make use of a free-body diagram of the tractor alone to determine the force between it and cart A. This exposes the coupling force \(\vec{T}\), which is our objective.

• $$\sum F_{x} = m_{system} a_{x}\; and\; \sum F_{x} = 820.0t,$$so $$820.0t = (650.0 + 250.0 + 150.0)a$$ $$a = 0.7809t \ldotp$$Since acceleration is a function of time, we can determine the velocity of the tractor by using a = \(\frac{dv}{dt}\) with the initial condition that v 0 = 0 at t = 0. We integrate from t = 0 to t = 3: $$\begin{split}dv & = adt \\ \int_{0}^{3} dv & = \int_{0}^{3.00} adt = \int_{0}^{3.00} 0.7809tdt \\ v & = 0.3905t^{2} \big]_{0}^{3.00} = 3.51\; m/s \ldotp \end{split}$$
• Refer to the free-body diagram in Figure \(\PageIndex{7}\)(b) $$\begin{split} \sum F_{x} & = m_{tractor} a_{x} \\ 820.0t - T & = m_{tractor} (0.7805)t \\ (820.0)(3.00) - T & = (650.0)(0.7805)(3.00) \\ T & = 938\; N \ldotp \end{split}$$

Since the force varies with time, we must use calculus to solve this problem. Notice how the total mass of the system was important in solving Figure \(\PageIndex{7}\)(a), whereas only the mass of the truck (since it supplied the force) was of use in Figure \(\PageIndex{7}\)(b).

Recall that v = \(\frac{ds}{dt}\) and a = \(\frac{dv}{dt}\). If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have dt = \(\frac{ds}{v}\) and dt = \(\frac{dv}{a}\). Now, equating these expressions, we have \(\frac{ds}{v}\) = \(\frac{dv}{a}\). We can rearrange this to obtain a ds = v dv.

Example 6.9: Motion of a Projectile Fired Vertically

A 10.0-kg mortar shell is fired vertically upward from the ground, with an initial velocity of 50.0 m/s (see Figure \(\PageIndex{8}\)). Determine the maximum height it will travel if atmospheric resistance is measured as F D = (0.0100 v 2 ) N, where v is the speed at any instant.

The known force on the mortar shell can be related to its acceleration using the equations of motion. Kinematics can then be used to relate the mortar shell’s acceleration to its position.

Initially, y 0 = 0 and v 0 = 50.0 m/s. At the maximum height y = h, v = 0. The free-body diagram shows F D to act downward, because it slows the upward motion of the mortar shell. Thus, we can write

\[\begin{split} \sum F_{y} & = ma_{y} \\ -F_{D} - w & = ma_{y} \\ -0.0100 v^{2} - 98.0 & = 10.0 a \\ a & = -0.00100 v^{2} - 9.80 \ldotp \end{split}\]

The acceleration depends on v and is therefore variable. Since a = f(v), we can relate a to v using the rearrangement described above,

\[a ds = v dv \ldotp\]

We replace ds with dy because we are dealing with the vertical direction,

\[\begin{split} ady & = vdv \\ (−0.00100v^{2} − 9.80)dy & = vdv \ldotp \end{split}\]

We now separate the variables (v’s and dv’s on one side; dy on the other):

\[\begin{split} \int_{0}^{h} dy & = \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} - 9.80)} \\ & = - \int_{50.0}^{0} \frac{vdv}{(-0.00100 v^{2} + 9.80)} \\ & = (-5 \times 10^{3}) \ln(0.00100v^{2} + 9.80) \Big|_{50.0}^{0} \ldotp \end{split}\]

Thus, h = 114 m.

Notice the need to apply calculus since the force is not constant, which also means that acceleration is not constant. To make matters worse, the force depends on v (not t), and so we must use the trick explained prior to the example. The answer for the height indicates a lower elevation if there were air resistance. We will deal with the effects of air resistance and other drag forces in greater detail in Drag Force and Terminal Speed .

Exercise 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?

Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

Newton's Second Law of Motion - Basic Problem

Please do provide a detailed step-by-step solution and a final answer.

• A truck with a mass of 3,500 kg, including the passengers, has an engine that produces a net horizontal force of 525 N. Assuming that there is no other forces involve in the motion, find the following:

a.) acceleration of the truck

b.) starting from rest, how long will it take for the truck to reach the velocity of 11.5 m/s?

2. An object with a mass m1 accelerates at 8.0 m/s2 when a force F is applied. A second object with a mass m2 accelerates at 4.0 m/s2 when the same force F is applied to it.

a.) Find the value of the ratio m2/m1.

b.) Find the acceleration of the combined mass under the action of force F.

1 Expert Answer

John B. answered • 10/20/21

Licensed Physics Instructor with Industry Experience

a) Since the engine is the only force acting, the net force is 525 N.

a=0.15 m/s 2

b) Using v=v 0 +at

11.5=0+(0.15)t

t=11.5/0.15

t=76.7 seconds

For the first mass F=m1(8)

For the second mass F=m2(4)

a) Therefore m2/m1 =(F/4)/(F/8) = 8/4 = 2

b) The combined mass is now m2 + m1 = F/8 + F/4 = 3F/8

Using F = (Combined Mass) x a

a= F/(3F/8)

a= 8/3 m/s 2 or 2.67 m/s 2

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Check Your Understanding

Answer: d = 1720 m

Answer: a = 8.10 m/s/s

Answers: d = 33.1 m and v f = 25.5 m/s

Answers: a = 11.2 m/s/s and d = 79.8 m

Answer: t = 1.29 s

Answers: a = 243 m/s/s

Answer: a = 0.712 m/s/s

Answer: d = 704 m

Answer: d = 28.6 m

Answer: v i = 7.17 m/s

Answer: v i = 5.03 m/s and hang time = 1.03 s (except for in sports commericals)

Answer: a = 1.62*10 5 m/s/s

Answer: d = 48.0 m

Answer: t = 8.69 s

Answer: a = -1.08*10^6 m/s/s

Answer: d = -57.0 m (57.0 meters deep) 

Answer: v i = 47.6 m/s

Answer: a = 2.86 m/s/s and t = 30. 8 s

Answer: a = 15.8 m/s/s

Answer: v i = 94.4 mi/hr

Solutions to Above Problems

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s 2 )*(32.8 s) 2

Return to Problem 1

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s) 2

110 m = (13.57 s 2 )*a

a = (110 m)/(13.57 s 2 )

a = 8.10 m/ s 2

Return to Problem 2

d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s 2 )*(2.60 s) 2

d = -33.1 m (- indicates direction)

v f = v i + a*t

v f = 0 + (-9.8 m/s 2 )*(2.60 s)

v f = -25.5 m/s (- indicates direction)

Return to Problem 3

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s 2

d = v i *t + 0.5*a*t 2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s 2 )*(2.47 s) 2

d = 45.7 m + 34.1 m

(Note: the d can also be calculated using the equation v f 2 = v i 2 + 2*a*d)

Return to Problem 4

-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s 2 )*(t) 2

-1.40 m = 0+ (-0.835 m/s 2 )*(t) 2

(-1.40 m)/(-0.835 m/s 2 ) = t 2

1.68 s 2 = t 2

Return to Problem 5

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s 2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s 2 )*(1.83 s) 2

d = 0 m + 406 m

Return to Problem 6

(7.10 m/s) 2 = (0 m/s) 2 + 2*(a)*(35.4 m)

50.4 m 2 /s 2 = (0 m/s) 2 + (70.8 m)*a

(50.4 m 2 /s 2 )/(70.8 m) = a

a = 0.712 m/s 2

Return to Problem 7

(65 m/s) 2 = (0 m/s) 2 + 2*(3 m/s 2 )*d

4225 m 2 /s 2 = (0 m/s) 2 + (6 m/s 2 )*d

(4225 m 2 /s 2 )/(6 m/s 2 ) = d

Return to Problem 8

d = (22.4 m/s + 0 m/s)/2 *2.55 s

d = (11.2 m/s)*2.55 s

Return to Problem 9

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(2.62 m)

0 m 2 /s 2 = v i 2 - 51.35 m 2 /s 2

51.35 m 2 /s 2 = v i 2

v i = 7.17 m/s

Return to Problem 10

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(1.29 m)

0 m 2 /s 2 = v i 2 - 25.28 m 2 /s 2

25.28 m 2 /s 2 = v i 2

v i = 5.03 m/s

To find hang time, find the time to the peak and then double it.

0 m/s = 5.03 m/s + (-9.8 m/s 2 )*t up

-5.03 m/s = (-9.8 m/s 2 )*t up

(-5.03 m/s)/(-9.8 m/s 2 ) = t up

t up = 0.513 s

hang time = 1.03 s

Return to Problem 11

(521 m/s) 2 = (0 m/s) 2 + 2*(a)*(0.840 m)

271441 m 2 /s 2 = (0 m/s) 2 + (1.68 m)*a

(271441 m 2 /s 2 )/(1.68 m) = a

a = 1.62*10 5 m /s 2

Return to Problem 12

• (NOTE: the time required to move to the peak of the trajectory is one-half the total hang time - 3.125 s.)

First use:  v f  = v i  + a*t

0 m/s = v i  + (-9.8  m/s 2 )*(3.13 s)

0 m/s = v i  - 30.7 m/s

v i  = 30.7 m/s  (30.674 m/s)

Now use:  v f 2  = v i 2  + 2*a*d

(0 m/s) 2  = (30.7 m/s) 2  + 2*(-9.8  m/s 2 )*(d)

0 m 2 /s 2  = (940 m 2 /s 2 ) + (-19.6  m/s 2 )*d

-940  m 2 /s 2  = (-19.6  m/s 2 )*d

(-940  m 2 /s 2 )/(-19.6  m/s 2 ) = d

Return to Problem 13

-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s 2 )*(t) 2

-370 m = 0+ (-4.9 m/s 2 )*(t) 2

(-370 m)/(-4.9 m/s 2 ) = t 2

75.5 s 2 = t 2

Return to Problem 14

(0 m/s) 2 = (367 m/s) 2 + 2*(a)*(0.0621 m)

0 m 2 /s 2 = (134689 m 2 /s 2 ) + (0.1242 m)*a

-134689 m 2 /s 2 = (0.1242 m)*a

(-134689 m 2 /s 2 )/(0.1242 m) = a

a = -1.08*10 6 m /s 2

(The - sign indicates that the bullet slowed down.)

Return to Problem 15

d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s 2 )*(3.41 s) 2

d = 0 m+ 0.5*(-9.8 m/s 2 )*(11.63 s 2 )

d = -57.0 m

(NOTE: the - sign indicates direction)

Return to Problem 16

(0 m/s) 2 = v i 2 + 2*(- 3.90 m/s 2 )*(290 m)

0 m 2 /s 2 = v i 2 - 2262 m 2 /s 2

2262 m 2 /s 2 = v i 2

v i = 47.6 m /s

Return to Problem 17

( 88.3 m/s) 2 = (0 m/s) 2 + 2*(a)*(1365 m)

7797 m 2 /s 2 = (0 m 2 /s 2 ) + (2730 m)*a

7797 m 2 /s 2 = (2730 m)*a

(7797 m 2 /s 2 )/(2730 m) = a

a = 2.86 m/s 2

88.3 m/s = 0 m/s + (2.86 m/s 2 )*t

(88.3 m/s)/(2.86 m/s 2 ) = t

t = 30. 8 s

Return to Problem 18

( 112 m/s) 2 = (0 m/s) 2 + 2*(a)*(398 m)

12544 m 2 /s 2 = 0 m 2 /s 2 + (796 m)*a

12544 m 2 /s 2 = (796 m)*a

(12544 m 2 /s 2 )/(796 m) = a

a = 15.8 m/s 2

Return to Problem 19

v f 2 = v i 2 + 2*a*d

(0 m/s) 2 = v i 2 + 2*(-9.8 m/s 2 )*(91.5 m)

0 m 2 /s 2 = v i 2 - 1793 m 2 /s 2

1793 m 2 /s 2 = v i 2

v i = 42.3 m/s

Now convert from m/s to mi/hr:

v i = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

v i = 94.4 mi/hr

Return to Problem 20