Question

A variable force $F = \left( {3x + 5} \right)N$ , acting on a body and if it is displaced from $x = 2m$ to $x = 4m$ , the work done by this force:

Answer 1

Hint:To solve this question, we must have a concept of variable force and how to deal with this question. As we know that when we have to calculate the work done simply, we have to integrate the equation. Or else in another way we have to use the graphical method where we have to calculate the area under the curve.

Complete step by step answer:
According to the question the function of force with respect to position is given as, $F = \left( {3x + 5} \right)N$ and also, we have given the interval i.e., $x = 2m$ to $x = 4m$ and we know to calculate the work done we have to integrate the following.As we know,
$W = \int {F \cdot ds} $
Where $F$ is the net force and $ds$ is the displacement.

So, integrating,
\[W = \int\limits_2^4 {\left( {3x + 5} \right)} \\
\Rightarrow W = \left( {\dfrac{{3{x^2}}}{2} + 5x} \right)_2^4 \\ \]
On further solving,
\[\Rightarrow W = \left( {\dfrac{{3 \times {4^2}}}{2} + 5 \times 4} \right) - \left( {\dfrac{{3 \times {2^2}}}{2} + 5 \times 2} \right) \\
\Rightarrow W = 44 - 16 \\
\therefore W = 28\,J \\ \]
Hence, the work done by the force acting on a body is $28J$.

Note:If the force exerted on the body is a constant force, then no integration is required to determine the force's work. Since the force will be constant, the integration will be based solely on displacement, which is the same as the body's net displacement. And don’t forget to write the unit at last and the unit of work is Joule.