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Question

Magnitude of work done by gravity during the motion of system is

A
0.8J
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B
1.6J
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C
0.4J
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D
4J
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Solution

The correct option is C 1.6J
Due to weight of block m, let spring gets elongated by distance x.
As one end of the rope is fixed , the block of mass m will go down by distance 2x.
Given:
Initial Velocity u=0 and at maximum extension
spring will come to rest for a moment so final Velocity v=0.
change in kinetic energy ΔKE=12mv212mu2
ΔKE=0

From work energy theorem , it is known

ΔKE=Wg+Wf+Ws, where Wg is work done by gravity,
Wf is work done by friction on block of mass 2m and Ws is work done by spring.
Wg=mg×2x=1×10×2x=20x

Ws=12kx2=0.5×100x2=50x2

Wf=μ2mgx=0.8×2×10x=16x

putting values in

ΔKE=Wg+Wf+Ws

0=20x50x216x

50x24x=0

x=0.08m

x=80cm

work done by gravity
Wg=mg×2x=1×10×2×0.08=1.6J

1652823_1742091_ans_e0d6c703aa6b453f87068c8d2a53963c.png

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