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Question: In which of the following situations is no work being done by the object exerting the force? A. An...

In which of the following situations is no work being done by the object exerting the force?
A. An elevator floor is exerting a force perpendicular to the bottom of a man’s shoes while the man is moving up.
B. A person is applying a normal force to a box to push it along a horizontal surface at a constant speed.
C. A person is holding a string that is exerting a force on a ball attached to the other end. The ball is moving in a circular path.
D. A person is slowly lowering a brick to the floor.
E. More than one of the preceding answers is correct.

Explanation

Solution

Hint :
-The Work done by an object is defined by the product of the force acting on it and the displacement of the object due to the force.
-Since both the force and displacement are the vector quantity and the work-done is a scalar quantity the product should be a Dot product.
-If the dot product is equal to zero, we can say there is no work done by an object.

Complete step by step answer:
A force is applied to an object at a certain angle and the object is displaced from its initial position.
Work done by the object is the dot product of the force and displacement since these twos are vector quantities and work done is a scalar quantity.
Let us explain this by the mathematical approach.
Let, F be the force vector acting on an object at an angle θ\theta , and S be the displacement vector.
The Work-done by the object
So we can write it as,
W=FScosθ\Rightarrow W = FS\cos \theta
Now if the value of cosθ\cos \theta is zero
So we can write the value of θ\theta is the odd multiple of π2\dfrac{\pi }{2}, the work done by the object should be zero.
Now we consider the given options:
A. An elevator floor is exerting a force perpendicular to the bottom of a man’s shoes while the
man is moving up:
In this case the angle between the displacement and the exerting force is 0{0^ \circ }.
Hence the work -done
W=FScosθW = FS\cos \theta
W=FS0\therefore W = FS \ne 0
Here we know that the value of cos0=1\cos 0 = 1
So, this is not the correct answer.
B. A person is applying a normal force to a box to push it along a horizontal surface at a constant speed:
In this case, the angle between the force acting on the box and the displacement of the box is 0{0^ \circ }. Hence the work-done,
W=FScosθW = FS\cos \theta
W=FS0\therefore W = FS \ne 0
Here the value of cos0=1\cos 0 = 1
So, this is not the correct answer.
C. A person is holding a string that is exerting a force on a ball attached to the other end. The ball is moving in a circular path:
When a person is holding a string that is exerting a force on a ball attached to the other end.
The ball is moving in a circular path and that’s why the angle becomes 90orπ2{90^ \circ }or\dfrac{\pi }{2}.
In this situation,
W=FScosθW = FS\cos \theta
W=FS=0\therefore W = FS = 0
Here the value of cosπ2=0\cos \dfrac{\pi }{2} = 0
So, here no work is being done by the ball.
Hence this is the correct answer.
D. A person is slowly lowering a brick to the floor: In this case, the angle between the force acting on the brick by the person and the displacement of the brick is 180{180^ \circ }. Hence the work is done,
W=FScosθW = FS\cos \theta
W=FS0\therefore W = - FS \ne 0
Here the value of cos0=1\cos 0 = - 1
So, this is not the correct answer.

Hence the correct option is (C).

Note: Work done can be zero when the displacement is zero. If a person is displaced by a force from its initial place but then back to the initial position again we can say there is no work being done by the person.
Also, if a person starts walking from a certain point of a circular path and back to this particular point covering the circle.
There is also no work being done by the person as the displacement is zero.