Chapter Review
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Chapter 7: Work and Kinetic Energy Chapter Review |
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Chapter Review
In chapters 5 and 6 you learned how to analyze mechanical situations by direct use of Newton's laws of motion. In this chapter you begin to learn about other concepts, which are consistent with Newton's laws, that can often make such analysis easier to perform. This is especially true when the number of applied forces is large, when they vary with distance or time, or when they are not accurately known.
7-1 Work Done by a Constant Force
When a force acts on an object that undergoes a displacement we say that the force does a certain amount of work on the object. This work can be positive, negative, or zero depending on how the direction of the force relates to the direction of the displacement.
If we consider the case in which a force F and the displacement d are in the same direction then we can consider the force as acting both through and with the displacement to its fullest extent. In this special case the work done by the force, W, will be positive and equal to the product of the magnitude of the force and the magnitude of the displacement: W = Fd. This result represents the maximum amount of positive work that the force f can do on the object. The opposite situation would be when the force and the displacement are in opposite directions (as is often the case with kinetic friction). Here, the force acts through and against the displacement to its fullest extent. In this latter case, the work done by the force will be negative W = -Fd. A third special case occurs when the direction of F is perpendicular to d (as is the case with the normal force on an object moving across a surface). When they are perpendicular, F neither works with nor against the displacement of the object and the work done by F on the object is zero.
In the most general cases of work done by a constant force, the force and displacement vectors are at arbitrary angles with each other. Consider the figure below
In part (a) we can see that F only partly acts with d and in part (b) it only partly acts against d. To determine the work done by F in such cases, we need to use only that part of F that acts with or against the displacement; that is, we need the component of F along the direction of d -- Fd. Based on the above diagrams, and what you know about vector components, try to convince yourself that Fd = F cos(q). In part (a) of the figure Fd will be positive and in part (b) it will be negative. The work done by F in this more general case, therefore, is given by
W = Fdd = (Fdcosq)d.
Notice that this general result for the work done by a constant force includes all three of the special cases discussed in the previous paragraph. As you can see in the above equation, the units of work must result from the product of the units of force and displacement. This combination 
, when applied to work, is called a joule (J).
It is useful to recognize that the work done can also be viewed as the component of the displacement in the direction of the force multiplied by the magnitude of the force. Algebraically, the difference only amounts to a regrouping of the factors involved
W = Fd(d cosq).
Try to convince yourself that the component of d along F really is dF = d cos(q). So, the important thing is that work only depends on the extent to which the force and the displacement act together (i.e., along the same direction).
When several forces act on an object while it is being displaced each force does work on the object according to the above discussion. The total work done on the object is the sum of all these contributions. An alternative way to approach this calculation is to first determine the total force acting on the object, Ftotal, and then calculate the work done by this force. Therefore, we have
,
where the total force, also called the net force, is given by the vector sum of all the forces 
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Physlet Illustration: Work Done Pulling a Block Horizontally | |
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| A block is pulled from rest by a constant force across a frictionless floor, as shown. The force on the box is represented by the black arrow. The velocity is given in m/s, and the distance grid is in meters. Adjust the mass (5 kg < m < 50 kg) and/or the pulling force (20 N< F < 200 N), and watch the graph of work done on the box vs. distance. How can you verify that the work done is calculated correctly? Start | |
Hints
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Example 7.1 Shopping for Groceries: A person pushes a 751 N shopping cart full of groceries down the aisle at the local store. The person applies a force of 112 N at an angle of 40.0 degrees below the horizontal. The cart is pushed the full length of a 15.5 m aisle at constant speed. Determine (a) the work done by the shopper, (b) the work done by gravity, (c) the work done by the normal force of the floor on the cart, and (d) the work done by various frictional forces during the cart's motion down the aisle.
Picture the Problem The picture shows the shopping cart, the force exerted by the shopper F, the weight mg, the normal force N and the displacement d.
Strategy To calculate the work done by the individual forces listed we need to identify the magnitude of each force and its direction relative to the displacement. The work done by frictional forces can be determined from the total work done on the cart.
Solution
Part (a)
| 1. From given information, the angle between F and d must be: | qF = 40.0o |
| 2. The work done by F then is: |
Part (b)
| 1. From the given situation the angle between the cart's weight and d is: | qg = 90.0o |
| 2. The work done by gravity is: | Wg = mgdcosqg = (751 N)(15.5 m)cos(90.0o) = 0J |
Part (c)
| 1. From given information, the angle between N and d must be: | qN = 90.0o |
| 2. The work done by N then is: | WN = NdcosqN = Ndcos(90.0o) = 0J |
Part (d)
| 1. Since the cart moves with constant speed the total force on the cart is: | Ftotal = ma = m(0 m/s2) = 0 |
| 2. The total work done on the cart then is: | |
| 3. Since the Wtotal = 0 the sum of the work done by each force must be zero: | |
| 4. Since Wg = WN = 0 this implies that: | Wfric = -WF = -1330J |
Insight Recognizing that both the normal force and the weight of the cart are perpendicular to the displacement, we could have immediately concluded that the work done by these forces is zero. However, the above approach shows that even if you don't make this recognition you can arrive at the correct result by performing the general calculation. Notice that the phrase "constant speed" in the statement of the problem was crucial to our ability to solve the problem. Be careful to pay attention to such phrases. Finally, take note that we were able to find the work done by friction forces without knowing anything about those forces. This point has caused many students to get stuck on similar problems. Keep in mind that there are many ways to get at information once you put all of the physics together.
Physlet Illustration: Work Done Pulling a Block at an Angle | |
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| A 20-kg block is being pulled by a rope across a frictionless floor, as shown. The rope makes an angle q with the horizontal. The velocity is given in m/s, and the distance grid is in meters. Adjust the angle (q < 90°) and/or the magnitude of the pulling force (F < 200 N), and monitor the work done on the block in the accompanying graph. Can you verify that the work displayed on the graph is calculated correctly? Start | |
Hints
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Exercise 7.2 Sliding Down: A 3.5 kg object slides 1.7 m down a ramp that is inclined at 35 degrees to the horizontal. If the coefficient of kinetic friction between the object and the ramp is 0.12, (a) how much work is done by gravity, and (b) how much work is done by kinetic friction?
Solution: The following information is given in the problem:
Given: m = 3.5 kg, d = 1.7 m, a = 35o, mk = 0.12; Find: (a) Wg, (b) WF
The diagram below shows the object on the ramp and the forces acting on it. Notice that the weight has been divided into its components parallel and perpendicular to the incline.
For part (a) we see that mgsin(a) is the component of the gravitational force in the direction of the displacement. So, we can immediately calculate the work done by gravity as
Wg = (mgsina)d = (3.5 kg)(9.81 m/s2)(1.7 m)sin(35o) = 33J.
For part (b) we need to determine the value of the force of kinetic friction. Since the object does not accelerate in the direction perpendicular to the incline, the two forces along that direction must cancel. This implies that
N = mgcosa.
The force of friction then is given by
.
Since fk always opposes the direction of motion, the angle between fk and d is 180o. This gives
There were two angles in this calculation, the angle of inclination of the ramp and the angle between the force and displacement. A common mistake is to get them confused and use the 35 degrees as the angle between the force and the displacement. Be careful about this situation. Also notice that unlike example 7.1, this time we determined the force of friction before calculating the work it did. Can you tell why it was more convenient to treat these two cases differently?
Physlet Illustration: The Work Done by Gravity on an Incline | |
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| A block slides down a frictionless incline, as shown. The incline makes an angle q with the horizontal. The speed is given in m/s, the times shown are in seconds, and the distance grid is in meters. Adjust the mass (100 g < m < 500 g) and/or the angle (5° < q < 30°), and watch the graph of work done on the block vs. distance it has travelled. How can you verify that the work done is calculated correctly? Start | |
Hints
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Practice Quiz
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