Question

A pile driver lifts a $250kg$weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground. A fall of $1.5m$drives the pipe in $35cm$. What is the average force exerted on the pipe?

Short answer

The average force exerted on the pipe is$10.5kN$.

Step-by-step solution

Content Introduction

The energy of closed systems remains constant according to the law of conservation of energy. It is possible to convert one form of energy into another. It is impossible to produce or destroy energy.

Content Explanation

According to conservation energy, $K_1+U_1=K_2+U_2$

Since, potential energy of driver can be converted into kinetic energy,

$U_1=K_2$

According to work energy theorem, the work done is equal to the change in kinetic energy. $W=K_2-K_1$

Substitute Fd₁ for W and $0$ for K₁.

$F=\frac{K_2}{d_2}$

Here, F is the force exerted on the pipe and d₂ is the distance of the pipe.

Substitute U₁ for K₂

$F=\frac{U_1}{d_2}$

The initial potential energy of the driver is,

$U_1=mg{d}_1$

Put the values and solve

$$\begin{gathered} F=\frac{(250kg)(9.81m/s^2)(1.5m)}{35cm\left({\displaystyle \frac{1m}{100cm}}\right)} \\ F=10.5kN \end{gathered}$$