Answer
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Hint:The force is calculated by multiplying the force by the amount of movement of an object, more precisely the product of the displacement and the component of force in the direction of displacement. Work is the mechanical manifestation of energy and has joule as its unit.
Formula Used: \[W=F.d.\cos \theta \]
Step by Step Solution:
The force acting on the load is only due to the earth’s gravitational field, or the weight of the body.
Let the mass of the body be \[m\text{ kg}\]. . The weight of the body would be \[mg\text{ N}\] .
Let the difference in height between the floor and the top of the building be \[\text{h metres}\] . This difference in heights is equivalent to the displacement of the body.
The person is applying force to raise the load and the load is being raised in the direction the force is being applied in, that is, the angle between the force and the displacement is \[\text{0}{}^\circ \] .
The work done in first case,
\[\begin{align}
& W=mgh\times \cos \text{0}{}^\circ \text{ J} \\
& \Rightarrow \text{W=mgh J (}\because \cos \text{0}{}^\circ =1) \\
\end{align}\]
Similarly, the work done in the second case
\[W'=mgh\text{ J}\]
We can see that \[W'=W\] , hence we can say that the work done in both the cases is the same
Option (C) is the correct answer.
Note:
We could have directly inferred the answer from the expression for work done; since there is no term for the time in the expression for work done, we can say that the work done is independent of the time taken and only depends upon the force and the displacement caused.
The power spent or developed depends on time, so we can say that the person who took \[10\text{ s}\] spent less power than the person who took \[\text{5 s}\] .
Formula Used: \[W=F.d.\cos \theta \]
Step by Step Solution:
The force acting on the load is only due to the earth’s gravitational field, or the weight of the body.
Let the mass of the body be \[m\text{ kg}\]. . The weight of the body would be \[mg\text{ N}\] .
Let the difference in height between the floor and the top of the building be \[\text{h metres}\] . This difference in heights is equivalent to the displacement of the body.
The person is applying force to raise the load and the load is being raised in the direction the force is being applied in, that is, the angle between the force and the displacement is \[\text{0}{}^\circ \] .
The work done in first case,
\[\begin{align}
& W=mgh\times \cos \text{0}{}^\circ \text{ J} \\
& \Rightarrow \text{W=mgh J (}\because \cos \text{0}{}^\circ =1) \\
\end{align}\]
Similarly, the work done in the second case
\[W'=mgh\text{ J}\]
We can see that \[W'=W\] , hence we can say that the work done in both the cases is the same
Option (C) is the correct answer.
Note:
We could have directly inferred the answer from the expression for work done; since there is no term for the time in the expression for work done, we can say that the work done is independent of the time taken and only depends upon the force and the displacement caused.
The power spent or developed depends on time, so we can say that the person who took \[10\text{ s}\] spent less power than the person who took \[\text{5 s}\] .
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