A mountain climber of mass 60\text{ kg} climbs a vertical height of 15\text{ m} in a time … — JAMB UTME

A mountain climber of mass $60\text{ kg}$ climbs a vertical height of $15\text{ m}$ in a time of $30\text{ s}$ . Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , calculate the average power expended by the climber in Watts.

Answer: 300 W

Answer

The average power expended by the climber is

300\text{ W}

The work done in lifting a mass

m

through vertical height

h

is given by

W = mgh = 60 \times 10 \times 15 = 9000\text{ J}

. The average power is the rate of doing work,

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

Step-by-Step Solution

Identify the given values and formula for work done against gravity.

Mass

m = 60\text{ kg}

, vertical displacement

h = 15\text{ m}

, acceleration due to gravity

g = 10\text{ m s}^{-2}

, and time

t = 30\text{ s}

. Work done formula:

W = mgh

The climber works against gravity to increase their potential energy by an amount equal to

mgh

Calculate the total work done.

W = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}

Multiplying force (

mg

) by vertical distance (

h

) yields work done in Joules.

Calculate the average power.

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

Power is defined as work done divided by the time interval (

P = \frac{W}{t}

Key Concept

Power as the rate of doing work against gravity

Estimated Time:45s

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