Question

A boy of mass \[50\text{ kg}\] runs up to a staircase of \[45\] steps in \[9\text{ s}\]. If the height of a step is \[15\text{ cm}\], find his power. (\[g=10\text{ m/}{{\text{s}}^{2}}\])

Answer 1

Hint: The boy possesses potential energy due to height above the ground, which is given by,
\[P.E.=mgh\]
Where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object from the ground.
Power of an object is the rate of energy per unit time.

Formula used:
The power P of an object is given by:
\[P=\dfrac{mgh}{t}\]
Where m is the mass of the object, g is the acceleration due to gravity, h is the height of the object from the ground, and t is the time in seconds.

Complete step by step solution:
Mass of the boy, \[m=50\text{ kg}\]
Time taken to run up the stairs, \[t=9\text{ s}\]
The staircase has \[45\] steps each of height \[15\text{ cm}\].
So, the height of the stairs, \[h=45\times 15\text{ cm}=675\text{ cm}=6.75\text{ m}\]
\[g=10\text{ m/}{{\text{s}}^{2}}\]
Substituting the value of mass, acceleration due to gravity, height and time in the power- formula:
\[\begin{align}
  & P=\dfrac{(50\text{ kg)(10 m/}{{\text{s}}^{2}})(6.75\text{ m)}}{9\text{ s}} \\
 & P=\dfrac{3375\text{ J}}{9\text{ s}} \\
 & P=375\text{ W} \\
\end{align}\]

Therefore the boy’s power is \[375\text{ W}\].

Additional information: The total energy of any object is conserved, that is, it can neither be created nor destroyed.
The potential energy of the boy at the base of the staircase is zero as his height from the ground is zero.

Note: The problem can be also solved by taking the work done by the boy in climbing the stairs.
Work done, \[W=F\cdot S\]
Where the force F on the body is the weight of the boy.
So, \[F=mg\]
The displacement S of the boy is equal to the height of the staircase.
Power of the boy is the rate of work done.
So, \[P=\dfrac{W}{t}=\dfrac{(mg)(h)}{t}=\dfrac{mgh}{t}\].