Question

when a load of 10kg is hung from the wire then extension of 2m is produced the work done by the restoring force

Answer 1

hope it useful to u. ok

Answer 2

Answer: 98 Joule is work done by restoring force.

Solution:

Given: Mass m = 10 kg  

    Extension x = 2 m

According to given question, the load of 10 kg is suspended from the wire which produces an extension of 2 m, the work done is calculated as followed.

As we know that work done by restoring force formula W=$\frac{1}{2} k x^{2}$,

After the extension when the load is suspended, equation formed for the system is mg = T  

T = k x

mg = k x

$k=\frac{m g}{x}$

Put the value of k in W

$\begin{aligned} W &=\frac{1}{2} k x^{2} \\ W &=\frac{1}{2}\left(\frac{m g}{x}\right) x^{2} \end{aligned}$

W=$\frac{1}{2} m g x$

W=$\frac{1}{2} \times 10 \times 9.8 \times 2$

W = 98 Joule .