Question

A boy is carrying a suitcase of $3kg$ mass on his back and moves $200m$ on a levelled road. If the value of $g$ be $10m{{s}^{-2}}$, the work done by the boy against the gravitational force will be:
$\begin{aligned} & A)6000J \\\ & B)0J \\\ & C)120J \\\ & D)800J \\\ \end{aligned}$

Answer 1

In Physics, work done on an object is dependent on the force applied on the object as well as the displacement caused on that object. For an object moving at a certain height from the earth’s surface on a levelled road, the vertical displacement caused due to the gravitational force on the object is equal to zero.

Formula used:
$W=Fs\cos \theta =mgh$

Complete answer:
Work is defined as the product of force and displacement. It is a vector quantity and has both magnitude as well as direction. When displacement caused is along the direction of the force applied, positive work is said to be done. On the other hand, when displacement caused is in the opposite direction of the applied force, negative work is said to be done. Mathematically, work is given by
$W=Fs\cos \theta$
where
$W$ is the work done on an object
$F$ is the force applied on the object
$s$ is the displacement caused due to the force
$\theta$ is the angle between the direction of force and the direction of displacement
Let this be equation 1.
When an object is held at a definite height from the surface of the earth, gravitational force acts on the object. When the object is released, this force causes a vertical displacement equal to the height of the object. Since gravitational force and vertical displacement in this case are in the same direction, work done by gravitational force is equal to the product of weight of the object and the height at which the object was held from the surface of the earth. This is given by
$W=mgh$
where
$W$ is the work done on an object by the gravitational force
$m$ is the mass of the object
$g$ is the acceleration due to gravity
$h$ is the vertical displacement or the height of the object
Let this be equation 2.
Coming to our question, we are told that a boy is carrying a suitcase of $3kg$ mass on his back and is moving a distance of $200m$ on a levelled road. If the value of $g$ be $10m{{s}^{-2}}$, we are required to determine the work done by the boy against the gravitational force.
Work done by the boy against the gravitational force is equal to the work done on the boy or the suitcase by the gravitational force.
Now, when the boy is moving, the magnitude of gravitational force on the suitcase is equal to the weight of the suitcase and the magnitude of vertical displacement is equal to zero, because he is walking on a levelled road. Using equation 2, work done on the suitcase by the gravitational force is equal to the product of weight of the suitcase and the vertical displacement. This is given by
$W=mgh=3kg\times 10m{{s}^{-2}}\times (0m)=0J$
Clearly, work done on the boy or the suitcase by the gravitational force is equal to $0J$. Therefore, the work done by the boy against the gravitational force is also equal to $0J$.

The correct option to be marked is $B$.

Note:
Students need to understand that both equation 1 and equation 2 are equal. For an object released from a certain height from the earth’s surface, work done by the gravitational force is given by
$W=Fs\cos \theta =(mg)h\cos (0)=mgh$
where
$W$ is the work done on the object by the gravitational force
$F=mg$, is the product of mass and acceleration, which is nothing but weight of the object
$s=h$, is the vertical displacement or height of the object
$\theta =0{}^\circ$ is the angle between force and vertical displacement of the object
Therefore, equation 1 can also be used to solve the given problem. Here, the direction of displacement is perpendicular to the direction of force and hence, the work done turns out to be zero because
$\cos \theta =\cos 90{}^\circ =0$