A 2100-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 5.00 m before coming into contact with the top of the beam. Then it drives the beam 12.0 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. a) m=2100 kg b) Xi=5.00m c) Xf=12.0 cm =.12m

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Answer:

   f = 878,080 N

Explanation:

mass of pile driver (m) = 2100 kg

distance of pile driver to steel beam (s) = 5 m

depth of steel driven (d) = 12 cm = 0.12 m

acceleration due to gravity (g0 = 9.8 m/s^{2}

calculate the average force exerted on the pile driver by the beam.

  • from work done = force x distance
  • work done = change in potential energy of the pile driver
  • equating the two equations above we have

               force x distance = m x g x (s - d)

              f x 0.12 = 2100 x 9.8 x (5- (-0.12))

              d = - 0.12 because the steel beam went down at we are taking its  

              initial position to be an origin point which is 0

              f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12

                   f = 878,080 N

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