Question

a body of mass M lying at rest on a smooth horizontal surface starts to move under the action of force F and gains velocity V after time T. the work done w by the force F

Answer 1

Given,
Mass of body = M
Foce acts on body = F
gains velocity after time T = V

We know, force = mass × acceleration [ according to Newton's 2nd law]
F = Ma
a = F/M , hence acceleration = F/M

Now, use formula,
S = ut + 1/2 at²
Because body starts to move from rest
so, initial velocity , u = 0
∴ S = 0 + 1/2(F/M)T²
S = (F/2M)T² hence, displacement = (F/2M)T²

Now, Workdone = force × displacement
= F × (F/2M)T²
= (F²/2M)T²

Answer 2

According to the Work Energy Theorem, the difference in the total energy of the body under observation is numerically equal to the work done by the same body. In this case, since the body does not move in the vertical direction, there is no change in the potential energy. Instead, the change in the motion in the horizontal direction results in a change in the kinetic energy of the body.

 

It is given that the body is initially at rest.

 

Initial kinetic energy = 0

Final kinetic energy = 0.5 x M x V^2

 

Difference in the kinetic energy = 0.5 x M x V^2

 

Hence, the work done by the body is 0.5 x M x V^2 Joules.