Answer to Question #150654 in Physics for damontae

Question

A boxer is working out on the heavy bag. On one of the boxer's punches the 41 kg heavy bag goes from hanging at rest to swinging at 6.59 m/s. The boxer's glove is in contact with the heavy bag for only 0.15 seconds.
With how much force, in N, does the boxer hit the heavy bag?

Answer 1

Impulse of the bag is:

$$\Delta p = mv-0 = mv$$

where $m = 41kg$ is the mass and $v = 6.59 m/s$ is the final speed. here stands for the initial momentum (in rest).

The average force is (according to the Newton's second law):

$$N = \dfrac{\Delta p}{\Delta t}$$

where $\Delta t = 0.15s$ is the time of interaction. Thus, obtain:

$$F = \dfrac{mv}{\Delta t}\\ F= \dfrac{ 41kg\cdot 6.59m/s }{0.15s} \approx 1801N$$

Answer. 1801 N.