Chapter 9, Problem 51A

Relative to the ground below, how many joules of PE does a 1000-N boulder have at the top of a 5-m ledge? If it falls, with how much KE will it strike the ground? What will be its speed on impact?

To determine

The potential energy at the top of the ledge, the kinetic energy at the ground and the velocity of the impact.

Answer to Problem 51A

  $5000\text{J, }5000\text{J}$ and $10\text{m/s}$

Explanation of Solution

Given:

The weight of the boulder, $W=1000\text{N}$

The height of the ledge, $h=5\text{m}$

Formula used:

As per law of conservation of energy, the energy of an isolated system cannot be destroyed or created. It always converts from one form to the another.

The gravitational potential energy is given by,

  $PE=mgh$

Where,

  $m=$ mass

  $g=$ acceleration due to gravity

  $h=$ height

And, the kinetic energy is given by,

  $KE=\frac{1}{2}m{v}^2$

Where,

  $v=$ velocity

The mass of the boulder can be determined as,

  $\begin{array}{c}m=\frac{W}{g}\\ =\frac{1000}{10}\\ =100\text{kg}\end{array}$

Then, the potential energy at the top of the ledge will be,

  $\begin{array}{c}PE=mgh\\ =Wh\\ =1000\left(5\right)\\ =5000\text{J}\end{array}$

Since the boulder was at rest at the top, hence its kinetic energy will be zero there. Similarly, the potential energy of the boulder will be zero at the floor. Now, considering the law of conservation of energy,

  $\begin{array}{l}PE=KE\\ KE=5000\text{J}\end{array}$

Now, using the above mentioned formula and the determined values,

  $\begin{array}{c}v=\sqrt{\frac{2\left(KE\right)}{m}}\\ =\sqrt{\frac{2\left(5000\right)}{100}}\\ =10\text{m/s}\end{array}$

Conclusion:

Hence, the potential energy at the top and the kinetic energy at the bottom will be same as $5000\text{J}$ while the velocity of the impact will be $10\text{m/s}$ .