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?

Question

asked Jul 14, 2020 94.2k views

1 Answer

Asym · Answer 1 · 2020-07-18T04:44:16+0000

Answer:

625 N

Step-by-step 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:

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