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 46 kg. If A and B are 4.5 and 0.5 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? UA = 0 m/s ng (a) vB = Number Units (b) vB = Number Units KEheavier - (C) KElighter Number Units

Explanation / Answer

a)

After descending a height of (h = hA - hB = 4.5 - 0.5 = 4.0m) ,

the KE of mass A = change in PE os mass A = mg*h

So, 0.5*mA*v^2 = mA*g*h

So, v = sqrt(2gh) = sqrt(2*9.8*4. 0) = 8.85m/s

b)

From the equation of v above we see that it does not depend on the mass of the particle

So, the velocity will be same as of mass A

So, v = 8.85m/s

c)

Ratio of KE = 0.5*mA*v^2 / 0.5*mB*v^2

= KE heavier/KE lighter=46/10=4.6

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