Question

a body of mass 5kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of body for 10 sec. The increase in kinetic energy?

Answer 1

Given, 
           mass(m) = 5 kg
           momentum(p) = mass*velocity = 10 kgm/s
           force(f) = 0.2 N
           time of force(t) = 10sec.

Solution,
             p= m*v =  10 kgm/s   
 (5 kg)*v = 10 kgm/s          
v = (10kgm/s)/5kg
= 2m/s
hence, initial velocity =2m/s (since it was already moving.)

f=ma
0.2N=(5kg)*a
a= (0.2N)/(5kg)
a= 0.04 m/s²

a= (final velocity-initial velocity)/change in time
a= Δv/Δt = (v-u)/Δt
0.04 m/s² = (v-2)/10sec
v= (0.04*10)+2
v = 2.4 m/s

change in kinetic energy = (mv²/2) - (mu²/2)
= (m/2) (v²-u²)
= (5/2) (2.4² - 2²) = 4.4 joule.

Answer 2

Given, 

          mass(m) = 5 kg

          momentum(p) = mass*velocity = 10 kgm/s

          force(f) = 0.2 N

          time of force(t) = 10sec.

Solution,

            p= m*v =  10 kgm/s   

 (5 kg)*v = 10 kgm/s          

v = (10kgm/s)/5kg

= 2m/s

hence, initial velocity =2m/s (since it was already moving.)

f=ma

0.2N=(5kg)*a

a= (0.2N)/(5kg)

a= 0.04 m/s²

a= (final velocity-initial velocity)/change in time

a= Δv/Δt = (v-u)/Δt

0.04 m/s² = (v-2)/10sec

v= (0.04*10)+2

v = 2.4 m/s

change in kinetic energy = (mv²/2) - (mu²/2)

= (m/2) (v²-u²)

= (5/2) (2.4² - 2²) = 4.4 joule.