Question

Forces of magnitudes $5$ and$3$ units acting in the directions $6i+2j+3k$ and $3i-2j+6k$respectively act on a particle which is displaced from the point $(2,2,–1)$ to $(4,3,1)$. Find the work done by the forces.

• A. $127/7units$
• B. $148/7units$
• C. $133/7units$
• D. $139/7units$

Answer 1

The correct option is B

$148/7units$

Step 1. Given data

The first force $F_1=5isalong(6i+2j+3k)$

The second force is $F_2=3isalong(3i-2j+6k)$

The particle is displaced from the point $\left(x_1,y_1,z_1\right)=(2,2,–1)$ to $\left(x_2,y_2,z_2\right)=(4,3,1)$

Step 2. work done by the forces is:

The two Forces on particle are

The unit vector is given by

$\hat{n}=\frac{\left(xi+yj+zk\right)}{\sqrt{x^2+y^2+z^2}}$

The force is then given by,

$\overrightarrow{F}=F\hat{n}$

$$\begin{gathered} \overrightarrow{F_1}=5\frac{\left(6i+2j+3k\right)}{\sqrt{6^2+2^2+3^2}} \\ \overrightarrow{F_2}=3\frac{(3i-2j+6k)}{\sqrt{3^2+2^2+6^2}} \end{gathered}$$

The resultant force is the sum of Force vectors

$$\begin{gathered} F_R=\overrightarrow{F_1}+\overrightarrow{F_2}=5\frac{\left(6i+2j+3k\right)}{\sqrt{6^2+2^2+3^2}}+3\frac{(3i+2j+6k)}{\sqrt{3^2+2^2+6^2}} \\ \Rightarrow F_R=5\frac{\left(6i+2j+3k\right)}{7}+3\frac{(3i-2j+6k)}{7} \\ \Rightarrow F_R=\frac{\left(30i+10j+15k\right)}{7}+\frac{(9i-6j+18k)}{7} \\ \Rightarrow F_R=\frac{\left(39i+4j+33k\right)}{7} \end{gathered}$$

The displacement vector of the particle is

$$\begin{gathered} \overrightarrow{d}=\left(x_2-x_1\right)i+\left(y_2-y_1\right)j+\left(z_2-z_1\right)k \\ \overrightarrow{d}=\left(4-2\right)i+\left(3-2\right)j+\left(1-\left(-1\right)\right)k \\ =2i+j+2k \end{gathered}$$

The work done is given by

$$\begin{gathered} W=F_R.d=\left(\frac{\left(39i+4j+33k\right)}{7}\right).\left(2i+j+2k\right) \\ =\frac{78+4+66}{7}=\frac{148}{7}units \end{gathered}$$

Hence, option B is correct