Question

How much force can a 2.5 kg sledge hammer excerpt on a nail if you can swing the hammer at 20 m/s and the hammer contacts the nail for 0.08 seconds?

Answer 1

Answer:

625 N

Explanation:

The impulse given to the nail is equal to the change in momentum of the hammer:

$I=F \Delta t=m \Delta v$

where

F is the force exerted by the hammer

$\Delta t=0.08 s$ is the time of contact

$m=2.5 kg$ is the mass

$\Delta v=20 m/s$ is the change in velocity of the hammer

Substituting the data and re-arranging the equation, we can find the force:

$F=\frac{m \Delta v}{\Delta t}=\frac{(2.5 kg)(20 m/s)}{0.08 s}=625 N$