The drawing shows two boxes resting on frictionless ramps. One box is relatively

Question

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 10 and 38 kg. If A and B are 3.0 and 1.0 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?

Explanation / Answer

At B both crates will have changed vertical position by a distance h of 3-1 =2 m
The change in kinetic energy will equal the change in potential energy
PE = KE
m*g*h =1/2*m*v^2
g*h = v^2/2
v = (2*g*h)
v = (2(9.81)2)
v = 6.264 m/s answers for part a & b

both boxes will be traveling at the same speed since they are attached. The mass and steepness of slope is not relevant.

ratio of the kinetic energies of heavy box to light box
ratio = KE1/KE2 =1/2*m1*v^2 /1/2*m2*v^2 = 38 / 10
ratio = 3.8:1
There is 3.8 times more kinetic energy in the 38 kg box than in the 10 kg box at the same velocity.

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