v = ? m/s At a construction site, a 60.0kg bucket of concrete hangs from a light

Question

v = ? m/s

At a construction site, a 60.0kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 80.0kg box on a horizontal roof (see the figure (Figure 1) ). The cable pulls horizontally on the box, and a 45.0kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving. Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.50m , from rest.

Explanation / Answer

mgh=1/2mv^2+1/2m'v^2+0.4*m'gx

So

60*9.8*2.5=0.5(80+60)v^2+0.4*80*9.8*2.5

So

v=3.13m/s

Check:

60g-T=60a

T-0.4*80g=80a

So

a=28g/140=1.96m/s^2

So

for s=2.5

v^2-u^2-2as

So

v=sqrt(2*1.96*2.5)=3.13m/s

Hence it is true

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