Question

A particle moves along the x-axis from $x = 0$ to $x = 1{\text{ }}m$ under the influence of a force given by $F = 3{x^2} + 2x - 10$. Work done in the process is
A) $+ 4\,J$
B) $- 4\,J$
C) $+ 8\,J$
D) $- 8\,J$

Answer 1

The work done by a particle when it gets displaced because of a force can be calculated as the dot product of the work and displacement. We will calculate the displacement of the particle and take its dot product with the force to calculate the work done by the force.

Formula used: In this solution, we will use the following formula:
$W = \int {F.dx}$ where $W$ is the work done by the particle, $F$ is the force acting on it, and $d$ is the displacement of the particle.

Complete step by step answer:
To calculate the work done, we must first find the displacement of the particle. The displacement of the particle depends only on its initial and final position. In this case, it goes from $x = 0$to $x = 1{\text{ }}m$.
Since we’ve been given that the force acting on the particle is, we can calculate the work done by the force using the relation:
$W = \int\limits_{x = 0}^{x = 1} {F.dx}$
Substituting the value of $F = 3{x^2} + 2x - 10$, we get
$W = \int\limits_{x = 0}^{x = 1} {\left( {3{x^2} + 2x - 10} \right)dx}$
On integrating, we can write
$W = \left. {{x^3} + {x^3} - 10x} \right|_{x = 0}^{x = 1}$
On placing the appropriate limits, we get
$W = - 8\,J$
Hence, the correct answer is option (D).

Note: The trick to solving this question is realizing that the work is done by the force only depends on the initial and the final position of the particle and not on the trajectory it takes since the displacement of the particle only depends on the initial and the final position. While calculating the work done here, since the force and the displacement are in the same direction that is towards the positive x-axis, we can replace the dot product with normal multiplication.