Question

A Force $20 + 10y$ acts on a particle in y-direction where F is in newton and y in a meter. Work done by this force to move the particle from $y = 0$ to $y = 1m$ is
A. $30J$
B. $5J$
C. $25J$
D. $20J$

Answer 1

Hint: We know work is the measure of energy expended in moving an object that force times displacement. No work is done if the object does not move, and the force is a physical quantity that has the ability to push or pull or accelerate a body. Also, for this work done is measured in joule.

Complete step by step answer:
We know that if a force F is acting on a particle and it displaces the particle by an infinitely small distance dx, then the work done by the force on the particle over the whole length x is given by,
$W = \int\limits_0^x {\overrightarrow F \cdot \overrightarrow {dx} } $
Here, W is the work done and it is equal to the dot product of force vector and displacement vector. Here integration is applied because to add every multiplication of force and displacement vector dx.
In the case of a variable force, integration is necessary to calculate the work done. Work done is equal to force and displacement.
By using integration;
We know that Work done by the particle along y direction is given by,
$W = \int\limits_0^1 {{\rm{F(y) dy}}} $
$ \Rightarrow W = \int\limits_0^1 {(20 + 10{\rm{y)}}} {\rm{dy}}$
$ \Rightarrow W = 20\int\limits_0^1 {{\rm{dy + 10}}\int\limits_0^1 {{\rm{ydy}}} } $
$ \Rightarrow W = 20[1 - 0] + 10\left[ {\dfrac{{{1^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]$
$\therefore W = 25\;{\rm{J}}$

Work done by this force to move the particle from $y = 0$ to $y = 1m$ is 25J. Hence, Option (C) is correct.

Additional information:
The same integration approach can also be applied to the work done by a constant force. This suggests that the integrating product of force and distance is the general way of determining the work done by a force on a moving body.

Note:
Work is not only a force product by displacement; it's the integration of the force over the displacement. If you have a force as a function of displacement and integrate it with respect to that displacement, you will get an expression for the work.