Question

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

If the mass of the paper is 0.003 kg, what force does the boxer exert on it? Include units.

Answer 1

Force exerted, F = 1.5 N

Step-by-step explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

`F=m* a`

`F=m* ((v-u)/(t))`

`F=0.003* ((30)/(0.06))`

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.