Question

Velocity of a body moving in a straight line is increased by applying a force 'F' for some distance in the direction of motion. Prove that increase in kinetic energy of the body is equal to the work done by the force on the body.

Explain with complete calculations & justifications.

Points : 10 ☺

Answer 1

let any body of mass m moves with constant velocity u in straight line , after sometimes apply force on the body , then body will be accelerate.
acceleration of body is F/m
now,
let displacement covered due to applying force F is S
now,
v^2 = u^2 +2aS
V^2 =u^2 +2(F/m). ----------------(1)

initial kinetic energy of body =1/2mu^2
final kinetic energy of body =1/2mv^2
from equation (1)
final K.E of body =1/2m(u^2+2FS/m)

now,
change in kinetic energy =∆K.E
∆K.E = K.Ef -K.Ei
=1/2mu^2-1/2m(u^2+2FS/m)
=F.S -------------(2)

again ,
if F force applying on a body .due to this body moves S distance linearly .
so,
workdone by force on body = F.Scos@
here body moves linearly hence,
@ =0°
so,
work done by force on body =F.Scos0°
=F.S -----------(3)
from equation (2) and (3)
hence,
change in kinetic energy =work done by force on body

Answer 2

let
mass of the body = m
initial velocity = u
final velocity = v
distance traveled before coming to rest = S
acceleration = a

force = mass x acceleration
F = m x a ---------------- (1)

work done = force x displacement
W = F X S ----------------(2)

from relation
v^2 = u^2 + 2aS

displacement S = (v^2 - u^2)/2a ---------------- (3)

substituting the value of F and S from eqns (1) and (3) in eqn (2) we get
W = m a x [(v^2 - u^2)/2a]
W = 1/2 m (v^2 - u^2)
W = 1/2 mv^2 - 1/2 mu^2

final KE = 1/2 mv^2
initial KE = 1/2 mu^2

W = final KE - initial KE

thus work done on the body = increase in KE