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

Question

The drawing shows two boxes resting on frictionless ramps. Onebox is relatively light and sits on a steep ramp. The other box isheavier and rests on a ramp that is less steep. The boxes arereleased from rest at A and allowed to slide down the ramps. Thetwo boxes have masses of 14 and 32 kg. If A and B are6.0 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 theheavier box to that of the lighter box at B?

Explanation / Answer

Both the boxes are at the same height, and are at rest, so: Their PE(potential energy)=mgha and their KE(kineticenergy)=0 (as velocity=0) Total energy=mgha At point B, some of this potential energy is converted tokinetic energy and the rest stays as potential energy. PE=mghb , KE=1/2mv2 Total energy =mghb + 1/2mv2 By conservation of energy, energy at A=energy at B mgha=mghb + 1/2mv2(canceling mass as it is common to all expressions) 2g(ha-hb)=v2 v=2g(ha-hb) This expression is common to both boxes and so their velocities areequal. v =2*9.8(6.0-1.0)       =9.899 m/s ( answer for aand b) KE=1/2mv2 KE1/KE2=m1/m2(as v1=v2) (where 1 is the heavier box)                 =32/14                 =2.2857

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