Question

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

Answer 1

Answer:

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.

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Answer 2

Answer:

Explanation:

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.