Question

What is the average power supplied by a $60.0kg$ person running up a flight of stairs a vertical distance of $4.0m$ in $4.2s$ ?

Answer 1

Power is the amount of energy converted per unit time and the SI unit of power is the watt. 1watt is equal to one joule per second and in a basic electric circuit power is equal to the product of voltage and current. The dimensional formula of power is $M{{L}^{2}}{{T}^{-3}}$ and power is a scalar quantity which has only magnitude.

Complete step-by-step solution:
Power is the amount of energy converted per unit time and work is the amount of energy transferred. In the bulb power is the rate at which electrical energy is converted into heat or light.
Power is the rate at which work is done with respect to time
$p=\dfrac{dw}{dt}$
Where W= work
T=time
At a distance z if we applied a force f then the work done is given as
$W=f\times z$
Then the power will be equal to

& p=\dfrac{dw}{dt}=\dfrac{d}{dt}(f\times z) \\\ & p=f\times \dfrac{dz}{dt} \\\ \end{aligned}$$ Velocity (v)=$\dfrac{dz}{dt}$ After substituting p=f$\times $ v Average power is defined as the ratio of total work done to the total time taken by the body and the unit of average power is kilowatt-hour. Work done against the gravity is given by W=mgh $\cdots \cdots (1)$ m=mass g=gravity which is equal to $9.8m{{s}^{-1}}$ h=height From the data we known the values of m=80kg h=4m and t=0.2s Substitute these values in equation (1) $w=60\times 9.8\times 4$ $w=2352J$ Average power $p=\dfrac{w}{t}$ We known $w=2352J$ $\begin{aligned} & p=\dfrac{2352}{4.2} \\\ & p=560watt \\\ \end{aligned}$ **Note:** Students in this problem work done against the gravity formula is used and the unit of time is second. Output power equation is different for different systems in case of motor the output power is the product of torque and angular frequency and in case of electric circuit the output power is product of voltage and current.