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下面小编为大家搜集整理了关于SAT2的物理练习题，希望对大家的备考有所帮助。
Practice Questions of Work, Energy, and Power
1. How much work does a person do in pushing a box with a force of 10 N over a distance of 4.0 m in the direction of the force?
(A)0.4 J
(B)4.0 J
(C)40 J
(D)400 J
(E)4000 J
2. A person pushes a 10 kg box at a constantvelocity over a distance of 4m. The coefficient of kinetic frictionbetween the box and the floor is0.3. How much work does the person doin pushing the box?
(A)12 J
(B)40 J
(C)75 J
(D)120 J
(E)400 J
3. How much work does the force of gravity doin pulling a 10 kg box down a30º inclined plane of length 8.0 m? Notethat sin 30 = cos 60 = 0.500and cos 30 = sin 60 = 0.866.
(A)40 J
(B)69 J
(C)400 J
(D)690 J
(E)800 J
4. How much work does a person do in pushing a box with a force of 20 N over a distance of 8.0 m in the direction of the force?
(A)1.6 J
(B)16 J
(C)160 J
(D)1600 J
(E)16000 J
5. The figure below is a force vs.displacement graph, showing the amountof force applied to an object bythree different people. Al appliesforce to the object for the first 4m of its displacement, Betty appliesforce from the 4 m point to the 6m point, and Chuck applies force fromthe 6 m point to the 8 m point.Which of the three does the most work onthe object?
(A)Al
(B)Betty
(C)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
(C)It is unchanged
(D)It is doubled
(E)It is quadrupled
7. A worker does 500 J of work on a 10 kgbox. If the box transfers 375 Jof heat to the floor through thefriction between the box and the floor,what is the velocity of the boxafter the work has been done on it?
(A)5 m/s
(B)10 m/s
(C)12.5 m/s
(D)50 m/s
(E)100 m/s
8. A person on the street wants to throw an 8kg book up to a personleaning out of a window 5 m above street level.With what velocity mustthe person throw the book so that it reachesthe person in the window?
(A)5 m/s
(B)8 m/s
(C)10 m/s
(D)40 m/s
(E)50 m/s
Questions 9 and 10 refer to aforklift lifting a crate of mass 100 kg ata constant velocity to aheight of 8 m over a time of 4 s. The forkliftthen holds the crate inplace for 20 s.
9. How much power does the forklift exert in lifting the crate?
(A)0 W
(B)2.0103 W
(C)3.2103 W
(D)2.0104 W
(E)3.2104 W
10. How much power does the forklift exert in holding the crate in place?
(A)0 W
(B)400 W
(C)1.6103 W
(D)4.0103 W
(E)1.6104 W
参考答案见下页
#p#参考答案与解析#e#
参考答案与解析
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
Thework done on the box is the force exerted multiplied by the box’sdisplacement. Since the box travels at a constant velocity, we knowthat the net force acting on the box is zero. That means that the forceof 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 bythe force of gravity is the dot product of the displacement of the boxand the force of gravity acting on the box. That means that we need tocalculate the component of the force of gravity that is parallel to theincline. This is mg sin 30 = (10 kg)(10 m/s2) sin 30. Thus, the work done is
4. C
This isthe same question as question 1. We were hoping that with differentnumbers and line spacing you wouldn’t notice. The test writers do thattoo sometimes.
5. C
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:
The amount of work Betty does is equal to the area of a triangle of length 2 and height 4:
The amount of work done by Chuck is equal to the area of a rectangle of length 2 and height 4:
J. We can conclude that Chuck did the most work.
Don’t be fooled by D: the force exerted by Al is in the opposite direction of the object’s displacement, so he does negative work on the object.
6. E
The formula for kinetic energy is KE =
mv2.Since the car’s kinetic energy is directly proportional to the squareof its velocity, doubling the velocity would mean quadrupling itskinetic energy.
7. A
The work-energy theorem tells us that the amount of work done on an objectis equal to the amount of kinetic energy it gains, and the amount ofwork done by an object is equal to the amount of kinetic energy itloses. The box gains 500 J of kinetic energy from the worker’s push,and loses 375 J of kinetic energy to friction, for a net gain of 125 J.Kinetic energy is related to velocity by the formula KE =
mv2, so we can get the answer by plugging numbers into this formula and solving for v:
8. C
When the book reaches the person in the window, it will have a gravitational potential energy of U = mgh.In order for the book to reach the window, then, it must leave thehands of the person at street level with at least that much kineticenergy. Kinetic energy is given by the formula KE = 1/2 mv2, so we can solve for v by making KE = U:
9. B
Power is ameasure of work divided by time. In turn, work is a measure of forcemultiplied by displacement. Since the crate is lifted with a constantvelocity, we know that the net force acting on it is zero, and so theforce exerted by the forklift must be equal and opposite to the weightof the crate, which is (100 kg)(10 m/s2) =
103 N. From this, we can calculate the power exerted by the forklift:
10. A
Poweris measured as work divided by time, and work is the dot product offorce and displacement. While the crate is being held in the air, it isnot displaced, so the displacement is zero. That means the forkliftdoes no work, and thus exerts no power.
以上就是小编为大家整理的关于Energy and Power的SAT2物理练习题，更多SAT干货敬请关注新航道SAT考试频道。
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