Question

4. 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) 50 J
(4) 25 J​

Answer 1

Answer:

-100

Explanation:

Given,

Given,m=10kg

Given,m=10kgx=2m

Given,m=10kgx=2mg=10m/s 2

Given,m=10kgx=2mg=10m/s 2By hook's law,

Given,m=10kgx=2mg=10m/s 2By hook's law,F=−kx

mg=kx

k=-mg/x

=− 10×10/2

k=−50N/m

k=−50N/mThe work done by the restoring force,

k=−50N/mThe work done by the restoring force,W= 1/2 kx2

W= 1/2×(−50)×2×2

W=−100J