Question

A 4.0 kilogram object moving west at 8.0 meters per second is subjected to a force to the east for 1.5 seconds, slowing it down to a speed of 2.0 meters per second. What was the net force that was applied?

Answer 1

This question relates to Newton's 2nd law.

Explanation:

Newton's 2nd law may be defined as `F= m(v-u)/t` (i.e. force applied equals the rate of change of momentum). Since acceleration, `a=(v-u)/t` , we get to the more commonly seen equation F=ma.
In the above, v= final velocity, u = initial velocity.
Applying to the above scenario:
F = [(3x2) -(3x8)]/1.5 = [6-24]1.5 =12N acting in the East direction