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?

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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?

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Answer 1 · 2016-02-28T21:11:36

This question relates to Newton's 2nd law.

Explanation:

Newton's 2nd law may be defined as $F = m \frac{v - u}{t}$ $F= m(v-u)/t$ (i.e. force applied equals the rate of change of momentum). Since acceleration, $a = \frac{v - u}{t}$ $a=(v-u)/t$ , 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

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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?

1 Answer

Explanation:

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