Question

14. A 0.135 kg ball is tossed upward from a height of 1.20 m above the ground with an initial speed of 15.0 m/s . What is the ball’s total mechanical energy at the moment it is tossed upward?(1 point)

Answer 1

To find the total mechanical energy of the ball at the moment it is tossed upward, we need to consider both its kinetic energy (KE) and its potential energy (PE) at that moment.

1. **Calculate the Kinetic Energy (KE)**:
The formula for kinetic energy is given by:

\[
KE = \frac{1}{2} mv^2
\]

where:
- \( m \) is the mass of the ball (0.135 kg)
- \( v \) is the initial speed (15.0 m/s)

Substituting in the values, we get:

\[
KE = \frac{1}{2} \times 0.135 \, \text{kg} \times (15.0 \, \text{m/s})^2
\]
\[
KE = \frac{1}{2} \times 0.135 \times 225
\]
\[
KE = \frac{1}{2} \times 30.375
\]
\[
KE = 15.1875 \, \text{J}
\]

2. **Calculate the Potential Energy (PE)**:
The formula for potential energy is given by:

\[
PE = mgh
\]

where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) is the height above the ground (1.20 m)

Substituting the values, we get:

\[
PE = 0.135 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.20 \, \text{m}
\]
\[
PE = 0.135 \times 9.81 \times 1.20
\]
\[
PE \approx 1.587 \, \text{J}
\]

3. **Calculate Total Mechanical Energy (E)**:
The total mechanical energy (E) is the sum of kinetic energy and potential energy:

\[
E = KE + PE
\]
\[
E = 15.1875 \, \text{J} + 1.587 \, \text{J}
\]
\[
E \approx 16.7745 \, \text{J}
\]

Therefore, the total mechanical energy of the ball at the moment it is tossed upward is approximately **16.77 J**.