Question

A 2,200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact with the top of the beam, and it drives the beam 15.8 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

Answer 1

here,

m = mass of the pile driver = 2200 kg

h = distance fallen to contact = 4.6 m

d = distance into the ground = 0.158 m

g = acceleration by gravity = 9.8 m/s^2

F = force exerted by I-beam on the pile driver = ?

The TOTAL distance the pile driver falls is

D = h + d

D = (4.6 m) + (0.158 m)

D = 4.758 m

The difference in potential energy of the pile driver between initial and final position is:

PE = m * g * D

PE = (2200 kg) * (9.8 m/s^2) * (4.758 m)

PE = 102687.2 J

That is also how much work that has to be done by the force exerted by the beam on the pile driver, so

PE = F*d

m * g * (h + d) = F * d

(102687.2 J) = F * (0.158 m)

F = 6.5 * 10^5 N

The pile driver is falling down. So to stop it, the I-beam must exert a force in the opposite directen.

F is upward