Question

A body constrained to move in y-direction is subjected to force given by $\vec F=(-2\vec i+15\vec j+6\vec k)N$. What is the work done by this force, in moving the body through a distance of $10\;m$ along the y-axis?

\[\begin{align}
  & A.20J \\
 & B.150J \\
 & C.160J \\
 & D.190J \\
\end{align}\]

Answer 1

Hint: We know that the work is defined as the product of force and displacement. Here we have a force which displaces the particle on the y direction. Hence to calculate the work, we must calculate the displacement of the particle and then use scalar multiplication to find the work done.
Formula used:
$W=F\cdot d$

Complete answer:
We know that work done is the scalar product of force $F$ and the displacement $d$. And is given as $W=F\cdot d=Fdcos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$.
Here, it is given that the $\vec F=(-2\vec i+15\vec j+6\vec k)N$, and the displacement is $10\;m$ along the y-axis.
Then we can write $\vec d=10\hat j$
Then, scalar multiplication of the $\vec F$ and$\vec d$ is given as
$W=(-2\vec i+15\vec j+6\vec k).(10\hat j)$
Since we know that, $i.j=0$ and $k.j=0$, as x,y, z axis are perpendicular to each other.
Then on substituting the values, we get
$W=15\times 10 cos 0$ ,since the angle made by the $\vec d$ is $0$ with respect to the y-axis.
Thus we get the total work done by the force $\vec F$ to move the object by $10\;m$ along the y-axis is $150J$

So, the correct answer is “Option B”.

Note:
The scalar multiplication can be used to multiply a scalar and a vector or two vector. If there are two vectors $\vec A$ and $\vec B$, then the scalar product is defined as $\vec A \cdot \vec B=|\vec A||\vec B|cos\theta$, where $\theta$ is the angle between the$\vec A$ and $\vec B$. Here we are integrating to get the work done, the student must know some basic integration to solve this sum. The scalar multiplication is also known as the dot product. Also, the resultant of scalar multiplication is always a scalar.
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