(A)Al
(B)Betty
?Chuck
(D)Al and Chuck do the same amount of work
(E)Betty and Chuck do the same amount of work
6. When a car’s speed doubles, what happens to its kinetic energy?
(A)It is quartered
(B)It is halved
?It is unchanged
(D)It is doubled
(E)It is quadrupled
7. A worker does 500 J of work on a 10 kg box. If the box transfers 375 J of heat to the floor through the friction between the box and the floor, what is the velocity of the box after the work has been done on it?
(A)5 m/s
(B)10 m/s
?12.5 m/s
(D)50 m/s
(E)100 m/s
8. A person on the street wants to throw an 8 kg book up to a person leaning out of a window 5 m above street level. With what velocity must the person throw the book so that it reaches the person in the window?
(A)5 m/s
(B)8 m/s
?10 m/s
(D)40 m/s
(E)50 m/s
Questions 9 and 10 refer to a forklift lifting a crate of mass 100 kg at a constant velocity to a height of 8 m over a time of 4 s. The forklift then holds the crate in place for 20 s.
9. How much power does the forklift exert in lifting the crate?
(A)0 W
(B)2.0 103 W
?3.2 103 W
(D)2.0 104 W
(E)3.2 104 W
10. How much power does the forklift exert in holding the crate in place?
(A)0 W
(B)400 W
?1.6 103 W
(D)4.0 103 W
(E)1.6 104 W
Explanations
1. C
When the force is exerted in the direction of motion, work is simply the product of force and displacement. The work done is (10 N)(4.0 m) = 40 J.
2. D
The work done on the box is the force exerted multiplied by the box’s displacement. Since the box travels at a constant velocity, we know that the net force acting on the box is zero. That means that the force of the person’s push is equal and opposite to the force of friction. The force of friction is given by , where is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the box, which is mg = (10 kg )(10 m/s2) = 100 N. With all this in mind, we can solve for the work done on the box:
3. C
The work done by the force of gravity is the dot product of the displacement of the box and the force of gravity acting on the box. That means that we need to calculate the component of the force of gravity that is parallel to the incline. This is mg sin 30 = (10 kg)(10 m/s2) sin 30. Thus, the work done is
4. C
This is the same question as question 1. We were hoping that with different numbers and line spacing you wouldn’t notice. The test writers do that too sometimes.
5. C
On a force vs. displacement graph, the amount of work done is the area between the graph and the x-axis. The work Al does is the area of the right triangle:英语作文
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