Question

A 2,200-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 3.40 m before coming into contact with the top of the beam, and it drives the beam 8.8 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.
magnitude N
direction

Answer 1

Answer:

The magnitude of Force is 8.58×10⁵N and direction is upwards

Explanation:

The work beam does on the pile driver is given by

W=(FCos180°)Δx= -F(0.088m)

From work energy theorem

$W_{nc}=(KE_{f}-KE_{i})+(PE_{f}-PE_{i})\\W_{nc}=1/2m(vf^{2}-vi^{2})+mg(yf-yi)$

Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m

So

$-F(0.088m)=1/2m(0-0)+2,200kg(9.81m/s^{2} )(-0.088m-3.40m)\\-F(0.088)=-75508.224\\F=75508.224/0.088\\F=8.58*10^{5} N$

The magnitude of Force is 8.58×10⁵N and direction is upwards