Question

A force at 4.0 N acts on a body of mass 2.0 Kg for 4 s. Assuming the body to be initially at rest, find.

(A) its velocity when the force stops acting,

(B) the distance covered in 10 S after the force starts acting.​

Answer 1

(A) When the force stops acting, it means force is zero newton, i.e., force stops acting means no force is acting on the body now.

Thus, force acting on the body = 0.

So, F = 0 N

m = 2.0 kg (mass remains constant)

F = m × a

or, 0 = 2 × a

or, a = 0

thus, acceleration is zero m/s^2 (metre per square second).

v = u + at

or, v = 0 + 0×t (since the body was initially at rest, it means its initial velocity, u is equal to zero)

or, v = 0

(B) When the force starts acting,

Acceleration = 2 m/s^2

time = 10 s

v = u + at = 0+20 = 20 m/s

v^2 - u^2 = 2.a.s

or, 20^2 - 0^2 = 2×2×s

or, 400 ÷ 4 = s

or, s = 100

thus, distance covered in 10 s = 100 m when the force starts acting

Answer 2

Answer:

u =0

F=4N

m=2 kg

t=4 s

(A) v=?

F=ma

4=2a

a=2m/s^2

a=v-u÷t

2=v-0÷4

v=8m/s

(B)

t=10s

s=ut +1÷2 at^2

s=(1÷2 ) 2×100

s=100 m

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