Question

a force of 10 Newton is applied on a object of mass 1 kg for 2 second which was initially at rest what is a work done on the object by the force

Answer 1

F=10 N
m=1 kg
t=2 s
and u=0 m/s
so, acceleration = F/m
= 10 m/s^2
therefore, s = 1/2 at^2
= 1/2 ×10×4
= 20 m
So, w= f×d
= 10 × 20
= 200 Joule

Answer 2

F=ma(from second law)

a=f/m

As force is 10N and mass is 1kg(given)

So a=10/1=10m/s^2

So, from 2nd equation of motion.

S=ut+1/2at^2

S=0×2+1/2×10×2^2(as the object is initially at rest so it's u=0)

So from here,

S=20m

Work done= force × distance travelled or displacement travelled

Work done=10×20=200J

So work done is 200J

Note: this question appears in stage 2 of NTSE, so choose the best solution.

HOPE IT HELPS YOU

REGARDS