Question

If under the action of a force ($4\widehat i + \widehat j + 3\widehat k$) N, a particle moves from position $\overrightarrow {{r_1}} = 3\widehat i + 2\widehat j - 6\widehat k$ to position $\overrightarrow {{r_2}} = 14\widehat i + 13\widehat j + 9\widehat k$, then what will be the work done?
${\text{A}}{\text{.}}$ 50 J
${\text{B}}{\text{.}}$ 75 J
${\text{C}}{\text{.}}$ 100 J
${\text{D}}{\text{.}}$ 175 J

Answer 1

Hint- Here, we will proceed by finding out the displacement covered by the given particle with the help of the initial and final position vectors. Finally, we will use the formula for the work done in vector form.

Formulas Used- $\overrightarrow d = \overrightarrow {{r_2}} - \overrightarrow {{r_1}}$, ${\text{W}} = \overrightarrow {\text{F}} .\overrightarrow d$ and $\overrightarrow x .\overrightarrow y = \left( {a\widehat i + b\widehat j + c\widehat k} \right).\left( {d\widehat i + e\widehat j + f\widehat k} \right) = ad + be + cf$.

Complete step-by-step solution -
Given, the uniform force (constant force) applied to a particle is $\overrightarrow {\text{F}} = 4\widehat i + \widehat j + 3\widehat k$ N
Initial position vector of the particle is $\overrightarrow {{r_1}} = 3\widehat i + 2\widehat j - 6\widehat k$
Final position vector of the particle is $\overrightarrow {{r_2}} = 14\widehat i + 13\widehat j + 9\widehat k$
As we know that the displacement vector $\overrightarrow d$ for any particle having initial position vector $\overrightarrow {{r_1}}$ and final position vector $\overrightarrow {{r_2}}$ is simply given by subtracting the initial position vector $\overrightarrow {{r_1}}$ from the final position vector $\overrightarrow {{r_2}}$
i.e., $\overrightarrow d = \overrightarrow {{r_2}} - \overrightarrow {{r_1}}$
By substituting the values of the initial position vector $\overrightarrow {{r_1}}$ and final position vector $\overrightarrow {{r_2}}$ for the given particle, we get

$$\Rightarrow \overrightarrow d = 14\widehat i + 13\widehat j + 9\widehat k - \left( {3\widehat i + 2\widehat j - 6\widehat k} \right) \\\ \Rightarrow \overrightarrow d = 14\widehat i + 13\widehat j + 9\widehat k - 3\widehat i - 2\widehat j + 6\widehat k \\\ \Rightarrow \overrightarrow d = 11\widehat i + 11\widehat j + 15\widehat k \\\$$

Also, we know that the work done W by any particle in vector form can be written as the dot product of the uniform force applied $\overrightarrow {\text{F}}$ on the particle and the displacement vector $\overrightarrow d$
${\text{W}} = \overrightarrow {\text{F}} .\overrightarrow d$
By substituting the values of the force applied and displacement occurred for the given particle, we get
$\Rightarrow {\text{W}} = \left( {4\widehat i + \widehat j + 3\widehat k} \right).\left( {11\widehat i + 11\widehat j + 15\widehat k} \right){\text{ }} \to {\text{(1)}}$
Also, the dot product of any two vectors $\overrightarrow x = a\widehat i + b\widehat j + c\widehat k$ and $\overrightarrow y = d\widehat i + e\widehat j + f\widehat k$ is given by

\Rightarrow \overrightarrow x .\overrightarrow y = ad + be + cf{\text{ }} \to {\text{(2)}} \\\ $$ From the above formula, it is clear that the dot product of any two vectors will be equal to the sum of the products of the corresponding coefficients of $$\widehat i,\widehat j,\widehat k$$ Using the above concept and formula given by equation (2) in equation (1), we get $ \Rightarrow {\text{W}} = 4 \times 11 + 1 \times 11 + 3 \times 15 \\\ \Rightarrow {\text{W}} = 44 + 11 + 45 \\\ \Rightarrow {\text{W}} = 100{\text{ J}} \\\ $ Therefore, the work done on the particle is 100 Joules. Hence, option C is correct. Note- A scalar quantity is the one having just magnitude whereas a vector quantity is the one with magnitude as well as direction. It is important to note that the dot product of any two vectors will always result in a scalar quantity. That’s why the work done is a scalar quantity because it is the dot product of two vectors i.e., force and displacement.