Question

Suppose a person does 2.75 × 10^{6} J of useful work in
7.75 h.

What is the average useful power output, in watts, of the person?

Working at this rate, how long, in seconds, will it take this person to lift 1950 kg of bricks 1.5 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output.)

Answer #1

Following information is given:

Work done by the person, W= 2.75×10^{6} J

Time taken by him, t = 7.75 h = 27900s

Formula for power is given

P = W/t

So, P = (2.75×10^{6} )/27900 = 98.566Watts

**The average output power is 98.566Watts**

**Now,** the work done to lift a load is the work
done against the gravity or equal to the gravitational potential
energy gained by the load.

Gravitational potential energy is given,

PE = m × g × h

Where, m is the mass of the load

And , g is acceleration due to gravity

And, h is the height to which the body is lifted off the ground

So, the work done to lift the given mass is :

W' = m*g*h = 1950 × (-9.8) × 1.5

W' = -28665 J

Negative sign indicates that work done is negative.

Now the work is being done at the rate of P, so, if t' time is taken to lift the load,then,

P = W'/t'

Or, t' = W'/ P

t' = 28665/ 98.566

t' = 290.82 s

**This person will take 290.82s to lift the
load.**

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