Question

15. A 0.09 kg beetle climbs vertically a distance of 0.35 m up a wall at a speed of 0.02 m/s. What is the beetle’s power output?(1 point)

Answer 1

To calculate the power output of the beetle, we can use the formula for power:

\[
P = \frac{W}{t}
\]

where \( P \) is power, \( W \) is work done, and \( t \) is time.

1. **Calculate the work done (W)**:
The work done against gravity can be calculated using the formula:

\[
W = mgh
\]

where:
- \( m = 0.09 \, \text{kg} \) (mass of the beetle)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.35 \, \text{m} \) (height climbed)

Substituting the values, we get:

\[
W = 0.09 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.35 \, \text{m} = 0.09 \times 9.81 \times 0.35
\]

Calculating this:

\[
W \approx 0.09 \times 9.81 \times 0.35 \approx 0.309 \, \text{J}
\]

2. **Calculate the time (t)**:
We can find the time it takes for the beetle to climb using its speed:

\[
t = \frac{h}{v}
\]

where \( v = 0.02 \, \text{m/s} \) is the speed of the beetle. Substituting the values:

\[
t = \frac{0.35 \, \text{m}}{0.02 \, \text{m/s}} = 17.5 \, \text{s}
\]

3. **Calculate the power (P)**:
Now that we have work and time, we can calculate power:

\[
P = \frac{W}{t} = \frac{0.309 \, \text{J}}{17.5 \, \text{s}} \approx 0.0176 \, \text{W}
\]

Thus, the beetle's power output is approximately **0.0176 watts**.