Question

Forces acting on a particle have magnitudes of 14,7 and 7N and act in the direction of vectors $6 \hat i +2 \hat j +3 \hat k$, $3 \hat i -2 \hat j +6 \hat k$, $2 \hat i -3 \hat j -6 \hat k$ respectively. The forces remain constant while the particle is displaced from A: (2, 1, -3) to B: (5, 1, 1). Find the work done. The coordinates are specified in meters.
A. 75J
B. 55J
C. 65J
D. 85J

Answer 1

Hint: To represent any vector quantity into its vector form, we need its magnitude and direction. The direction of any vector can be found out by a unit vector in its direction. A unit vector is a vector having magnitude unity and is used only to represent the direction of a vector. A unit vector is represented by $\hat n$.

Formula used:
$\hat n= \dfrac{\vec v}{|v|}$, $W = \int{ \vec F .ds}$, $s= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Where v is any vector quantity.

Complete step by step answer:
Here, we are given three forces’ magnitudes and some vectors in its direction. To find $\vec F$, we use : $\hat n= \dfrac{\vec v}{|v|}$, by replacing v by F.
Hence $\vec F = |F|.\hat n$
Now, given vectors are: $6 \hat i +2 \hat j +3 \hat k$, $3 \hat i -2 \hat j +6 \hat k$, $2 \hat i -3 \hat j -6 \hat k$
Let’s find their unit vectors by dividing them with their magnitudes;
$n_1 = \dfrac{6 \hat I + 2\hat j +3\hat k}{\sqrt{6^2+2^2+3^2}}$=$\dfrac{6 \hat I + 2\hat j +3\hat k}{7}$
$n_2 = \dfrac{3 \hat I -2\hat j +6\hat k}{\sqrt{3^2+2^2+6^2}}$=$\dfrac{3 \hat i - 2\hat j +6\hat k}{7}$
$n_3 = \dfrac{2 \hat I -3\hat j -6\hat k}{\sqrt{2^2+3^2+6^2}}$=$\dfrac{2 \hat i -3\hat j -6\hat k}{7}$
Hence to get $\vec F$, multiplying magnitudes of forces by their respective unit vectors:
$F_1=14 \times \dfrac{6 \hat i + 2\hat j +3\hat k}{7}$$= \ {12 \hat i + 4\hat j +6\hat k}$
$F_2=7 \times \dfrac{3 \hat i - 2\hat j +6\hat k}{7}$= $3 \hat i -2 \hat j +6 \hat k$
$F_3=7 \times \dfrac{2 \hat I -3\hat j -6\hat k}{7}$= $2 \hat i -3 \hat j -6 \hat k$
Now, in equation $W = \int{ \vec F .ds}$ ,$\vec F$is the net force.
Which is $\vec F_{net} = \vec F_{1}+\vec F_{2}+\vec F_{3}$=$\ {12 \hat i + 4\hat j +6\hat k}+3 \hat i -2 \hat j +6 \hat k+2 \hat i -3 \hat j -6 \hat k$
Hence $\vec F_{net} = 17 \hat i - \hat j +6 \hat k$
Or $|\vec F_{net}|= \sqrt{17^2+1^2+6^2}$=18.055N
Now, to find displacement vector $\vec s=(x_2-x_1) \hat i + (y_2-y_1) \hat j +(z_2-z_1) \hat k$
Putting the values of coordinates to find displacement vector:
$\vec s = (5-2) \hat i + (1-1) \hat j + (1-(-3)) \hat k = 3 \hat i + 0 \hat j + 4\hat k$
Now, as force is constant, work done will be the dot product of$\vec F \ and \ \vec s$.
So, $W = \vec F . \vec s = 17\times 3 + 0 + 6\times 4 = 51 +24 = 75J$

So, the correct answer is “Option A”.

 Note:
Since here force is constant as well as displacement is not the function of path, hence we could directly take the dot product of force and displacement. If in case, force were variable or displacement
Not constant, that the only tool we have is integration, by writing ds=dx+dy+dz.