Question

When a load of 10 kg is hung from the wire, then extension of 2 m is produced. Then work done by restoring force is (1) -200 J (2) -100 J (3) 80J (4) -25 J​

Answer 1

Answer:

The correct option is B)

Given,

m=10kg

x=2m

g=10m/s

2

By hook's law,

F=−kx

mg=−kx

k=−

x

mg

​

=−

2

10×10

​

k=−50N/m

The work done by the restoring force,

W=

2

1

​

kx

2

W=

2

1

​

×(−50)×2×2

W=−100J