- #1
MNWO
- 10
- 0
Hi guys, so I've been given a physics question which is black magic to me, so I thought i will ask you for help :) Here is the question:
A pile driver hammer of mass 170kg falls freely through a distance of 6m to strike a pile of mass 410kg and drives it 75mm into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.
This is what i have done so far:
m1=170kg
m2=410kg
D1=6m
D2=75mm
PE= m x g x h
PE= 170 x 9.81 x 6
PE=10006.2J
Velocity=2 x g x s
(2) x (9.81) x (6-0)=117.72 square root=10.84 m/s
KE= ½ x mass x velocity
KE=1/2 x 170 x 10.84*
KE=9987.97J
Work= change in kinetic energy + change in potential energy + work due to friction
½ ( 170kg) ( 10.84*-0) + (150kg)(9.81m/s*) (6-0)
=18816.97N without work due to friction
Now I don't know how to work out: work due to friction and how to calculate the resistance of the ground. This is what I was able to do myself and now I am stuck.
A pile driver hammer of mass 170kg falls freely through a distance of 6m to strike a pile of mass 410kg and drives it 75mm into the ground. The hammer does not rebound when driving the pile. Determine the average resistance of the ground.
This is what i have done so far:
m1=170kg
m2=410kg
D1=6m
D2=75mm
PE= m x g x h
PE= 170 x 9.81 x 6
PE=10006.2J
Velocity=2 x g x s
(2) x (9.81) x (6-0)=117.72 square root=10.84 m/s
KE= ½ x mass x velocity
KE=1/2 x 170 x 10.84*
KE=9987.97J
Work= change in kinetic energy + change in potential energy + work due to friction
½ ( 170kg) ( 10.84*-0) + (150kg)(9.81m/s*) (6-0)
=18816.97N without work due to friction
Now I don't know how to work out: work due to friction and how to calculate the resistance of the ground. This is what I was able to do myself and now I am stuck.