Question

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The box experiences a friction force of 30 N as it slows to a stop. How long does it take the box to stop?

Answer 1

The time taken to stop the box equals 1.33 seconds.

Step-by-step explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

`F=mass* acceleration\\\\\therefore acceleration=(Force)/(mass)`

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

`acceleration=(30)/(5)=6m/s^(2)`

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

`v=u+at\\\\0=8-6* t\\\\\therefore t=(8)/(6)=4/3seconds`

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.