Question

A 2.0- kg box is moving at constant speed of 2.0 m/s on a frictionless horizontal floor. You apply a horizontal force with a magnitude of 4.0 N to make it stop. How long should you apply this force?

Answer 1

Answer:

1 second

Explanation:

Impluse is the product of force and time. Specifically the force over a time interval is know as impulse. From the second law of motion we get impulse being equal to the change in momentum.

u = Initial velocity = 2 m/s

v = Final velocity = 0

t = Time taken

F = Force applied = 4 N

$Ft=m(v-u)\\\Rightarrow 4t=2(2-0)\\\Rightarrow 4t=4\\\Rightarrow t=\frac{4}{4}\\\Rightarrow t=1\ s$

Time the force has to be applied is 1 second.