Question

the average resisting force that must act on a 5 kg mass to reduce its speed from 65cm/s to 15cm/s in 0.2s is

Answer 1

F*dt = mv - mu
F = m (v - u) /dt = 5*(65-15)/0.2*100=5*50/20=12.5N

Answer 2

Answer:

Force, F = -12.5 N

Explanation:

Mass of the object, m = 5 kg

Initial speed of the object, u = 65 cm/s = 0.65 m/s

Final speed of the object, v = 15 cm/s = 0.15 m/s

Time, t = 0.2 s

Let F is the average resisting force. It can be calculated using second law of motion. It is given by :

$F=m\times a$

$F=m\times \dfrac{v-u}{t}$

$F=5\times \dfrac{0.15-0.65}{0.2}$

F = -12.5 N

So, the magnitude of the resisting force is 12.5 N. Hence, this is the required solution.