Question

A 1,800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.00 m before coming into contact with the top of the beam, and it drives the beam 14.4 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

Answer:

average force = 385,140 N

Explanation:

from the question we are given the following

mass (m) = 1800 kg

distance of fall (d) = 3 m

driven distance (l) = 14.4 cm = 0.144 m

acceleration due to gravity (g) = 9.8 m/s^{2}

work done = average force x driven distance.....equation 1

and

work done = change in kinetic energy + change in potential energy

work done = (0.5 x m x (v^{2} - u^{2})) + (m x g x (-d-l))

• Initial velocity (u) and final velocity (v) are zero because the pile driver is it rest before it moves to hit the pile and after hitting the pile.
• The changes in length for the potential energy are negative because the pile moves downward

we now have work done = (m x g x (-d-l))...equation 2

now equating the two equations for work done we have

average force x driven distance = (m x g x (-d-l))

average force x 0.144 = 1800 x 9.8 x (-3-0.144)

average force = (1800 x 9.8 x (-3-0.144)) ÷ 0.144

average force = 385,140 N