Question

A body of mass 5 kg is moving with a momentum of 10 kg metre per second a force of 0.2 newton act on it in the direction of motion of body of for 10 seconds increase in kinetic energy is

Answer 1

Answer:

Explanation:

Mass(m) = 5 kg (Given)

Momentum(p) = mass × velocity = 10 kgm/s (Given)

Force(f) = 0.2 N (Given)

Time of force(t) = 10sec (Given)

= p= m × v =  10 kgm/s   

= 5 ×v = 10 kgm/s          

v = (10)/5

= 2m/s

Thus, the initial velocity =2m/s

f = ma

0.2N = 5 × a

a = (0.2N)/(5)

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)/10

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