Question

A force acts on a body of mass 20 kg initially at rest.The force acts for 0.1 s and is then withdrawn and the body moves with the velocity of 2 m/s.Find magnitude of force.


i) 300 N

ii) 400 N

iii) 45 N

iv) 2000 N

​

Answer 1

Explanation:

Mass, m=2kg

Initial velocity, u=0

Final velocity, v=2m/s

Change in velocity, Δv=v−u=2m/s

Time interval, Δt=0.1s

Acceleration, a=

Δt

Δv

=

0.1

2

=20m/s

2

Force, F=ma=2kg×20m/s

2

=40N

HERE IS YOUR ANSWER. HOPES IT HLPS YOU. MARK ME AS A BRAINLIST. THANKUUUUUUU

Answer 2

$\sf \: mass \: = \: 20 \: kg \\ \sf \: initial \: velocity(u) \: = 0 \: {ms}^{ - 1} \\ \sf \: final \: velocity (v)\: = 2 \: {ms}^{ - 1} \\ \sf time \: = \: 0.1 \: s \\ \: \sf \: Acceleration(a) \: = ?$

$\sf \: a \: = \frac{v - u}{t} \\ \\ \sf \: a \: = \: \frac{2 - 0}{0.1} \\ \\ \sf \: a \: = \: \frac{2 \times 10}{1} \\ \\ {\sf{ \implies \: a = \: 20 \: {ms}^{ - 2} }}$

therefore,

Force = m×a

=> 40 N (ans)