Question

A 95-kilogram student climbs 4.0 meters up a rope in 3.0 seconds. What is the power
output of the student?
(a) 1.3 ´102 W
(b) 1.2 ´103 W
(c) 3.8 ´102 W
(d) 3.7 ´103 W
(e) n.o.t.a.

Answer 1

Answer on Question #40968, Physics, Mechanics

A 95-kilogram student climbs 4.0 meters up a rope in 3.0 seconds. What is the power output of the student?

(a) $1.3 \cdot 10^{2} \mathrm{~W}$

(b) $1.2 \cdot 10^{3} \mathrm{~W}$

(c) $3.8 \cdot 10^{2} \mathrm{~W}$

(d) $3.7 \cdot 10^{3} \mathrm{~W}$

(e) n.o.t.a.

Solution:

Given:

$$\begin{array}{l} m = 95 kg, \\ h = 4.0 \, m, \\ t = 3.0 \, s, \\ P = ? \\ \end{array}$$

Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.

$$Power = \frac{Work}{time}$$

In our case work is increasing the potential energy of body

$$W = mgh$$

Thus,

$$P = \frac{mgh}{t} = \frac{95 \cdot 9.81 \cdot 4}{3} = 1242.6 = 1.2 \cdot 10^{3} \, \text{Watts}$$

Answer. (b) $1.2 \cdot 10^{3} \, \text{W}$