Question

A weight lifter lifted a load of $200{\text{kgf}}$ to a height of $2 \cdot 5{\text{m}}$ in $5{\text{s}}$. Calculate the power developed by him. Take $g = 10{\text{Nk}}{{\text{g}}^{ - 1}}$ .
A) $10{\text{W}}$
B) $1{\text{W}}$
C) $1000{\text{W}}$
D) $100{\text{W}}$

Answer 1

Hint:As the weight lifter lifts the load off the ground to the given height, work will be done by him. The work done depends on the applied force and the displacement it causes. The power generated refers to the amount of work done in the given time.

Formulas used:
The work done by an applied force is given by, $W = F \cdot d$ where $F$ is the applied force and $d$ is the displacement it causes.
The power generated is given by, $P = \dfrac{W}{t}$ where $W$ is the work done by the applied force and $t$ is the time taken.

Complete step by step answer.
Step 1: List the parameters mentioned in the question.
The problem at hand involves a weightlifter lifting a load off the ground to some height in some time.
The weight of the load is given to be $W = 200{\text{kgf}}$ .
The height at which it is raised corresponds to the displacement of the load and it is given to be $d = 2 \cdot 5{\text{m}}$ .
The time taken to lift the load is $t = 5{\text{s}}$ .
Step 2: Express the work done by the force applied to lift the load.
The work done by the applied force can be expressed as $W = F \cdot d$ ---------- (1)
where $F$ is the applied force and $d$ is the displacement it causes.
Here the weight of the load corresponds to the applied force so $W = F = 200{\text{kgf}}$ .
Substituting for $F = 2000N$ and $d = 2 \cdot 5{\text{m}}$ in equation (1) we get, $W = 2000 \times 2 \cdot 5 = 5000{\text{J}}$
Thus the work done by the weight lifter is $W = 5000{\text{J}}$ .
Step 3: Express the relation for the power developed in the weight lifter.
The power generated can be expressed as $P = \dfrac{W}{t}$ --------- (2)
Substituting for $W = 5000{\text{J}}$ and $t = 5{\text{s}}$ in equation (2) we get, $P = \dfrac{{5000}}{5} = 1000{\text{W}}$
Thus the power developed is $P = 1000{\text{W}}$ .

So the correct option is C.

Note:The load refers to the force due to gravity or the weight. But here it is expressed in units of kgf. So conversion of the unit from kgf to Newtons is done using the conversion formula$W{\text{(in kgf)}} \times g = W{\text{(in N)}}$. On substituting we get, $200{\text{kgf}} \times 10{\text{Nk}}{{\text{g}}^{ - 1}} = 2000{\text{N}}$. This is then substituted in equation (1) to find the work done by the weight lifter.