Question

A body of mass 5kg initially at rest is moved by a horizhorizontal force of2n on smooth horizontal surface the work done by force in 10s is

Answer 1

Principle : Work Energy Theorem + Kinematic
F=ma
F/m= a= 0.4m/s²
initially At Rest means u=0
in 10 secs
v= u+ 0.4*10
v= 4m/s
Now Work done is change in Kinetic Energy
so work done = KEf - KEi
but KEi = 0 as initially the body was at rest and by the formula we know KEi= 1/2*m*u
so work done is = KEf
= 1/2 * 5kg*16m²/s²
=40kgm²/s²
= 40J