Question

How is work done by a force measured when the force:
(A) Is in the direction of displacement.
(B) Is it an angle to the direction of displacement.

Answer 1

Hint: We know that work done by a force on an object is given by the product of force and displacement i.e., \[W = F \times S\] and when the force is applied on the object at an angle with displacement then the work done by the object is given by the product of force, displacement, and angle. i.e., \[W = F \times S\cos \theta \] .

Complete step by step answer
We know that work is defined as the amount of energy transferred to or from an object via the application of force along with a displacement. We can say that when a force is applied on an object and it displaces by some distance then the work done by the force is equal to the product of the force applied and the displacement in the object.
The SI unit of work is Joule; it is represented by ‘J’.
The work done by the force can be expressed as \[W = F \times S\] . Here, W is representing work, F is representing force applied on the object whereas S is representing the displacement in the object.
1 Joule is equal to the product of one newton and 1 meter.
In the first case, we have to find the work done by a force when force is in the direction of displacement.
In this case, the work done by the force on the object is given by-product of force applied on the object and the displacement in the object. i.e., \[W = F \times S\] .
Similarly, in the second case, we have to find the work done by the force when the force is applied to the object at an angle to the direction of displacement.
In this case, the work done by the force on the object is given by-product of force applied, displacement, and the angle at which the force is applied to the displacement. It can be expressed as \[W = F \times S\cos \theta \] .

Additional information
It is noted that work is a scalar quantity. Another unit of work is foot-pound. The one-foot pound is equal to 1.35 joules. The Joule is the SI unit of work.

Note
One can do a mistake by writing the second expression of work done as \[W = F \times S\sin \theta \] , but it is noted that the work done by the object is given by-product of force applied, displacement, and the angle at which the force is applied to the displacement. It can be expressed as \[W = F \times S\cos \theta \] .