Question

Q3. A force of 6 N acts on a body of mass of 1.5 kg for 2 s. Assuming the
body to be initially at rest, find –
a. its velocity when the force stops acting.
b. the distance covered in 5 s after the force starts acting.

Answer 1

Answer:

acceleration of body for 2 sec

a = 6/1.5 = 4 m/s2

velocity after 2 sec

v = a*t = 4*2 = 8 m/s

Distance coververd in 5 sec

d = ( ½*4*22) + (8*3) = 8+24 = 32 m

Hope it clears.

Answer 2

Answer:

acceleration of body for 2 sec

a = 6/1.5 = 4 m/s2

velocity after 2 sec

v = a*t = 4*2 = 8 m/s

Distance coververd in 5 sec

d = ( ½*4*22) + (8*3) = 8+24 = 32 m

Hope it clears.