Question

Body of mass 100kg is dropped from height 125m in 5sec find work done in 5th second

Answer 1

so you are saying that
m= 100 kg
s= 125 m.
t= 5s. (total time taken to reach the ground)
 the work done just before reaching the ground is same as the potential energy at the initial position.
= mgh
= 100×10×125=125000 joule,( work done is same as the energy)

Answer 2

Given:-

$m=100kg$

$h=125 m$

$u=0\frac{m}{s}$

$a=g=10 \frac{m}{s^{2} }$

$t=5s$

What to find:-

Work done

Procedure:-

Let us use newton's third law of motion

$2as=v^{2} -u^{2}$

$2(10)(125)=v^2$

$2500=v^2$

$v=50 \frac{m}{s}$

$W_n_e_t=\frac{1}{2}m \int\limits^v_u {x} \, dv^{2}$

        $=\frac{1}{2} mv^{2} -\frac{1}{2} mu^{2}$

       $=50(2500)-0$  ( $\frac{100}{2}=50$ and u=0 causing $\frac{1}{2} mv^2$ term to get cancelled)  

       =125,000 J