Question

Your answer A boxer punches a sheet of paper in midair, and brings it from rest up to a speed of 40 m/s in 0.05 s. If the mass of the paper is 0.01 kg, what is the force of impact on the paper?

Answer 1

Answer:

Force of impact, F = 8 N

Explanation:

It is given that,

Mass of the paper, m = 0.01 kg

Initial speed of the paper, u = 0

Final speed of the paper, v = 40 m/s

Time taken, t = 0.05 s

Firstly finding the acceleration of the paper using the equation of kinematics as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{40\ m/s-0}{0.05\ s}$

$a=800\ m/s^2$

Let F is the force of impact on the paper. It is equal to the product of mass and the acceleration as :

$F=m\times a$

$F=0.01\ kg\times 800\ m/s^2$

F = 8 N

So, the force of impact on the paper is 8 N. Hence, this is the required solution.