Chapter 8: Q101P (page 211)

A 0.63 kg ball thrown directly upward with an initial speed of 14 m/s reaches a maximum height of 8.1 m . What is the change in the mechanical energy of the ball–Earth system during the ascent of the ball to that maximum height?

Short Answer

Expert verified

The change in mechanical energy of the ball is - 12 J .

Step by step solution

Given data:

Mass of the ball, m = 0.63 kg
Initial speed, $v_{0} = 14 m / s$
Maximum height. $h = 8.1 m$

To understand the concept:

Define the loss in mechanical energy oftheball earth system by taking the difference between initial mechanical energy and final mechanical energy.

Formula:

The change in energy is,

$∆ E = m g h - \frac{1}{2} m v_{0}^{2}$

Here, $∆ E$ is the change in energy, m is the mass, g is the acceleration due to gravity having a value 9.8 $m / s^{2}$ , h is the maximum height, and $v_{0}$ is the initial speed.

Calculate the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height:

Consideringthepoint of throwing as a reference point, you have

$∆ E = m g h - \frac{1}{2} m v_{0}^{2} = ((0.63 kg) \times (9.8 m / s^{2}) + (8.1 m)) - (\frac{1}{2} \times (0.63 kg) \times {(14 m / s)}^{2}) = (50.00094 J) - (61.74 J) \approx - 12 J$