A 2,100-kg pile driver is used to drive a steel beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam, and it drives the beam 12.0 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.

858375 N of force is exerted on the pile driver by the beam.

Explanation:

mass of the pile driver m = 2100 kg

distance of fall = 5 m

distance through which the beam is driven = 12 cm = 0.12 m

weight of the pile driver = mg

where g = acceleration due to gravity = 9.81 m/s^2

weight of pile driver = 2100 x 9.81 = 20601 N

work done by gravity in bringing the pile driver done the 5 m height is

work = weight x distance = 20601 x 5 = 103005 J

This work by gravity is also used to do work in driving the beam into the 12 cm depth.

The force exerted by the beam on the pile driver will be proportional to the force used to do the work in driving the beam through the 12 cm depth.

equating the works, we have

103005 = F x 0.12

F = 103005/0.12 = 858375 N of force

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.