Question

A particle moves from a point $(2\hat i + 5\hat j)$ to $(4\hat j + 3\hat k)$ when a force of $(4\hat i + 3\hat j)\,N$ is applied. How much work has been done by the force?
A) 2J
B) -8J
C) -11J
D) 5J

Answer 1

Hint
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. Calculate the displacement of the particle and take its dot product with the force to calculate the work done by the force.

Formula used:
$\Rightarrow W = F.d$ 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 so it can be calculated as the difference between the final and the initial position of the particle:
$\Rightarrow \vec d = (4\hat j + 3\hat k) - (2\hat i + 5\hat j)$
$\Rightarrow - 2\hat i - \hat j + 3\hat k$
Since we’ve been given that the force acting on the particle is $(4\hat i + 3\hat j)\,N$ , we can calculate the work done by the force using the relation:
$\Rightarrow W = F.d$
Substituting the values of $F = (4\hat i + 3\hat j)\,N$ and $\vec d = - 2\hat i - \hat j + 3\hat k$ , we can calculate the dot product as:
$\Rightarrow W = (4\hat i + 3\hat j)\,.( - 2\hat i - \hat j + 3\hat k)$
$\Rightarrow - 8 - 3 = - 11\,J$
Hence the work done by the force is equal to $- 11\,J$ which corresponds to option (C).

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 dot product, we must be careful in multiplying and adding components that have the same direction only.