Question

An $\mathbf{820}\mathbf{N}$ Marine in basic training climbs a $\mathbf{12}.\mathbf{0}\mathbf{m}$ vertical rope at a constant speed in 8.00 s. What is his power output?

Short answer

The Marine’s power output is $1.23\text{kW}$.

Step-by-step solution

Power

The total amount of work done by the system in the unit time interval is called power. It is given as follows:

$\mathbf{P}\mathbf{=}\frac{\mathbf{W}}{\mathbf{t}}$

Here,

$\mathbf{W}\mathbf{=}$Work done

$\mathbf{t}\mathbf{=}$ Time in second

Given Data

Weight of Marine:$F=820\text{N}$

Length of rope climbed: $L=12.0\text{m}$

Time taken: $t=8.00\text{s}$

Explanation

The Marine must exert an $820\text{N}$ upward force, opposite the gravitational force, to lift his body at a constant speed. So, the total work done should be equal to the increase in the potential energy of Marine. Thus, Marine’s power output is given as follows:

$$\begin{gathered} P=\frac{W}{t} \\ =\frac{U_f-U_i}{t} \\ =\frac{mgh-0}{t} \\ =\frac{mgh}{t} \end{gathered}$$

Substitute the numerical values:

$$\begin{gathered} P=\frac{\left(820\text{N}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(12.0\text{m}\right)}{8.00s} \\ =1230\text{W} \\ =1.23\text{kW} \end{gathered}$$

The total power output is $1.23\text{kW}$.