The weight of on object on the moon is one-sixth its weight on the earth. A body
ID: 1558813 • Letter: T
Question
The weight of on object on the moon is one-sixth its weight on the earth. A body moving with a given speed on the moon has kinetic energy equal to ___ it would have if it were moving at the same speed on the earth. A) the kinetic energy B) 1/36 the kinetic energy C) 1/6 the kinetic energy D) 6 times the kinetic energy E) 36 times the kinetic energy A light rope rum through two functionless pulleys of negligible mass. A mass, m, is hung from one of the pulley and a force, F, is applied to one end of the rope so that the mass moves at a constant speed. If the force it applied over a distance d the distance the mass moves is A) 0 B) 1/2 d C) 1/2 d D) 4 E) 2d The graph shows the speed of a lights truck of mass 4000 kg coasting on neutral down a level straight road. The truck is slowing down due to air resistance and rolling friction. Estimate the work done per second by the retarding force at t = 25 s A) -20 kJ b) 40 kJ C) -60 kJ D) -80 kJ E) -100 kJ Consider two engines. The larger is rated at 2 W and the smaller at 1 W. The smaller one can a certain quantity of work in 2 h. The larger can do twice as much work in a time of A) 30 min B) 1 h C) 2 D) 4 h E) 1.4 h Two skiers at the same lace at the top of the hill, and finish at the same place at the bottom. Skier A takes a straight, smooth route to the finish whereas Skier B takes a curvy bumpy route to the finish. If you assume that friction is negligible, which of the following statements is true? A) Skier A has the same speed as Skier B at the finish. B. Skier B has the greater speed at the finish. C) Skier A has the greater speed at the finish. A 40-kg girl, standing at rest on the ice, gives a 60 kg boy, who is also standing at rest on the ice, a shove. After the shove, the boy is moving backward at 2 m/s. Ignore friction The girls speed is A) zero B) 1.3 m/s C) 2 m D) 3 m/s E) 6 m/s Is you take derivative of the kinetic energy of a particle with respect to its velocity you get A) force BH) momentum C) acceleration D) mass E) potential energy.Explanation / Answer
1)
Ans: A) the kinetic energy.
Kinetic energy = 1/2 mv2, which is independent of g, weight depends on g but not mass.
2)
Ans: D) d
The mass has not accelerated up since constant velocity is maintained. The net work done must be conserved
F x d = mg x
Since F = mg, d = x
3)
Initial velocity, vi = 20 m/s
Velocity at t = 5s , vf = 15 m/s
Work done = change in kinetic energy
= 1/2m [vf2 - vi2]
= 0.5 x 4000 x [225 - 400]
= -350000 J
Work done per second = -350000/5
= -70000 J
= - 70 kJ
4)
Work done by smaller engine in 2 h = Power x time
= 1W x 2hr = 2 W.hr
Work that has to be done by the larger engine = 2 x work done by smaller
= 2 x 2 W.hr = 4 W.hr
Time taken to do this work = Work/Power
= 4 W.hr / 2W = 2 hr
5)
Initial kinetic energy of the skier A = 0
Initial potential energy = mgh
Consider that the kinetic energy at the bottom = 1/2 mv2
Change in kinetic energy = 1/2 mv2 - 0 = mgh
v = SQRT[2gh]
The final velocity depends only on the height between the initial and final points, which is same for both the skiers. So they have the same speed at the bottom.
6)
Initial momentum = Final momentum
0 = sum of momentum of girl's and boy's momentum
Consider that the girl has a velocity of v m/s
0 = 60 x 2 - 40 x v
v = 120/40 = 3 m/s
7)
d/dv [1/2 mv2]
= 1/2 m d/dv[v2]
= 1/2 m x 2v
= mv, which is the momentum
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