Question

A stone of mass 5 kg moving with a velocity of 0.2 m/ s is stopped by a player in 0.5 s. What is the force applied by the man to stop the stone.

Answer 1

Answer:

-2N

Explanation:

m = 5kg

u = 0.2 $\frac{m}{s}$

v = 0 $\frac{m}{s}$

t = 0.5sec

finding a by; v = u + at

$0 = 0.2 + 0.5a\\\\-0.5a = 0.2\\\\a = \frac{-0.2}{0.5}\\\\a = -0.4 \frac{m}{s^2}$

by F = ma

$F = 5 * (-0.4)\\\\F = -2N$