Question

A 2 200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.20 m before coming into contact with the top of the beam, and it drives the beam 15.4 cm farther into the ground before coming 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:

$F = 6.10 \times 10^5 N$

Explanation:

Here we know that pile will go down under gravity for d = 4.20 m before coming in contact with beam

Now after being in contact with the beam the pile further go down by L = 15.4 cm.

Finally it will come to rest.

So here we can say that work done by gravity + work done by beam on the pile must be equal to change in its kinetic energy

So here we have

$mg(L + d) - F(L) = 0$

$F = \frac{mg(L + d)}{L}$

$F = \frac{2200\times 9.81(4.20 + 0.154)}{0.154}$

$F = 6.10 \times 10^5 N$