Chapter 4, Problem 58R

To determine

To calculate: Speed of the boulder just before it hits the ground.

Answer to Problem 58R

Speed of the boulder just before it hits the ground is $\underset{\_}{62.60\text{ m/s}}$ .

Explanation of Solution

Given:

Mass = $\text{2500 kg}$

Height = $\text{200 m }$

Formula used:

Gravitational potential energy is given by,

$GPE=mgh$

where $m$ is mass, $g$ ( $\text{9}\text{.8 N/kg}$ ) is gravity and $h$ is height.

Kinetic energy is given by,

$KE=\frac{1}{2}m{v}^2$

where $m$ is mass and $v$ is speed.

Calculation:

Gravitational potential energy at the top is calculated as:

$\begin{array}{l}GPE=mgh\\ =(2500\text{ kg)(9}\text{.8 N/kg)(200 m)}\\ =4900000\text{ J}\end{array}$

When the boulder is at rest, the kinetic energy will be zero. Since mechanical energy is the sum of kinetic and potential energy in a system,

  $\begin{array}{l}\text{Mechanical energy = kinetic energy + potential energy}\\ =0+4900000\text{ J}\\ \text{ = 4900000 J}\end{array}$

When the boulder falls, the gravitational potential energy is converted to kinetic energy. But the sum total of kinetic energy and potential energy (mechanical energy) remains the same. Just before the boulder hits the ground, the kinetic energy is greatest and gravitational potential energy is smallest. Considering that the sum remains constant, this is only possible if kinetic energy is $\text{4900000 J}$ and gravitational potential energy is zero.

Speed of the boulder just before it hits the ground is calculated as:

$\begin{array}{l}KE=\frac{1}{2}m{v}^2\\ v=\sqrt{\frac{2KE}{m}}\\ =\sqrt{\frac{2\times 4900000\text{ J}}{2500\text{ kg}}}\\ =62.60\text{ m/s}\end{array}$

Conclusion:

Speed of the boulder just before it hits the ground is $\underset{\_}{62.60\text{ m/s}}$ .