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A pile driver lifts a 450 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground. A fall of 1.5 m drives the pipe in 45 cm.
What is the average force exerted on the pipe?

Respuesta :

syed514
the gravitational potential energy becomes work driving the pipe 

m g h = f d 

450 * g * 1.5 = f * .45

Answer:

F = 14700N

Explanation:

Hello! Let's solve this!

We have the following data:

vi = 0m / s

dy = 1.5m

m = 450kg

dx = 0.45m

First we need to know the speed of the impeller.

We can use the following formula

[tex]vf^{2}=vi^{2}+2*g*dy[/tex]

We get the vf

[tex]vf=\sqrt{2*g*dy}[/tex]

[tex]vf=\sqrt{2*9.8m/s^{2}*1.5m }[/tex]

vf = 4.42m / s

Then the kinetic energy is equal to the work done.

[tex]KE=\frac{1}{2}*v^{2} *m[/tex]

KE=1/2*[tex]5.42m/s^{2}[/tex]*450kg

We cleared KE

KE = W = 6615J

From the working formula we will clear the force exerted.

W = F * dx

F = W / dx = 6615J / 0.45m

F = 14700N

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