Question

7. A force of 4.0 N acts on a body of mass 2.0 kg for
4.0 s. Assuming the body to be initially at rest, find
fa) its velocity when the force stops acting,
b) the distance covered in s after the force starts
acting.​

Answer 1

Explanation:

f=ma. f=m(v/t). 4=2(v/4). v=8m/s. v^2-u^2=2as a=2m/s^2. s=64/4=16m

Answer 2

As we know that,

F = ma

4 = 2 × a

a = 4/2

a = 2 ms-²

Now,

a). v = u + at

v = 0 + 2 × 4

v = 8 ms-¹

b). S = ut + 1/2at²

S = 0 + 1/2 × 2 × 4 × 4

S = 16 m