## A body at rest is acted upon by a force for 20 seconds. The force is then withdrawn and the body moves a distance of 60m in the next 5 secon

Question

A body at rest is acted upon by a force for 20 seconds. The force is then withdrawn and the body moves a distance of 60m in the next 5 seconds. If the mass of the body is 10 kg, calculate the magnitude of the force.​

in progress 0
3 years 2021-08-30T15:52:30+00:00 2 Answers 12 views 0

6 N

Explanation:

Given:

v₀ = 0 m/s

v = 60 m / 5 s = 12 m/s

t = 20 s

Find: a

a = Δv / Δt

a = (12 m/s − 0 m/s) / 20 s

a = 0.6 m/s²

F = ma

F = (10 kg) (0.6 m/s²)

F = 6 N

6N

Explanation:

From the principle of conservation of momentum.

The momentum initiated by the Force F in 20sec is the impulse and is given as;

F×t= F×20 = 20F

Similarly the mass 10kg acquires a momentum of m×v; where m is mass =10kg and v is the velocity of the body.

The velocity of the body, V = Distance/ time =60/5=12m/s

Hence the momentum is;

10×12= 120Ns

By conservation of momentum;

20F =120

F = 120/20 = 6N