Chapter 6, Problem 76GP

To determine

a. The work done by gravity on the elevator

b. The speed of the elevator just before striking the spring

c. Theamount by which the spring compresses

Answer to Problem 76GP

Solution:

a. The work done by gravity is $258,352J.$

b. The speed of the elevator is $23.5m/s.$

c. The compression of the spring is approximately $2.43m$ .

Explanation of Solution

Given:

The mass of the elevator is $m=925\text{kg}$ .

The height of the elevator when the $h=28.5\text{m}$

The spring constant of the spring $k=8.00\times {10}^4\frac{\text{N}}{\text{m}}$

Formula used

The work done by gravity is $W_g=mgh$ .

For part b, the kinetic energy of the spring is equal to the work done by gravity.

Thus,

  $\begin{array}{l}\frac{1}{2}m{v}^2=mgh\\ v=2gh\end{array}$

For part c,the kinetic energy of the elevator is equal to the spring energy and the potential energy of the spring extension.

Calculations

a

The work done by gravity on the elevator is,

  $W_g=mgh$

  $\begin{array}{l}W=925\times 9.8\times 28.5\text{J}\\ W=258,352.5\text{J}\end{array}$

b

The speed of the elevator just before striking the spring is,

  $\begin{array}{l}{v}^2=2gh\\ v=\sqrt{2\times 9.8\times 28.5}\frac{\text{m}}{\text{s}}\\ v=23.6\frac{\text{m}}{\text{s}}\end{array}$

c

  $\begin{array}{l}\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2+mgx\\ 258,352.5=\frac{1}{2}\times 8.00\times {10}^4{x}^2+925\times 9.8x\\ x=9.72\text{m},-2.43\text{m}\end{array}$

Compressed length $=2.43\text{m}$

Conclusion:

a. The work done by gravity is $258,352J.$

b. The speed of the elevator is $23.5m/s.$

c. The compression of the spring is approximately $2.43m$ .