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Question

A block of mass 10 kg pulled by force 100 N. It covers a distance 500 m in 10s. From initial point this motion is observed by two observers A and B as shown in fig.



Find out work done by force F in 10 seconds w.r.t. A and B respectively.

A
5000 J and 5000 J
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B
5000 J and 0
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C
4000 J and 0
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D
0 and 6000 J
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Solution

The correct option is C 4000 J and 0

Distance covered by block =S
Distance covered by A=SA
SA=uA×t=10×10=100 m
Work done with respect to A is
(WF)w.r.tA=F[SSA]=100[500100]
(WF)w.r.tA=4000 J
Distance covered by B=SB
Now,
By applying 2nd law of motion,
we get
SB=uBt+12aBt2
SB=0+12×10×102
SB=500 m
Workdone with respect to B is
(WF)w.r.tB=F[SSB]=100(500500)
(WF)w.r.tB=0

Answer 4000 J,0

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