Question

A 40-kilogram mass is moving across a horizontal surface at 5.0 m/s. What is the magnitude of the net force required to bring the mass to a stop in 8.0 seconds? 

Answer 1

Given:

$m=\rm 40\: kg$

$v_i=\rm 5.0\: m/s$

$v_f=\rm 0\: m/s$

$\Delta t=\rm 8\: s$

The Newton's second law says

$$F=ma=m\frac{\Delta v}{\Delta t}$$

$$F=\rm 40*\frac{5.0-0}{8}=25\: N$$