Work Done By A Constant Force - Practice Questions & MCQ

Two Months Preparation Plan for JEE Main and JEE Advanced 2024

Quick Facts

Nature of Work Done is considered one of the most asked concept.
11 Questions around this concept.

Solve by difficulty

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that

its velocity is constant

its acceleration is constant

its kinetic energy is constant

it moves in a straight line.

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A force $\vec{F}= \left ( 5\hat{i} +3\hat{j}+2\hat{k}\right )N$ is applied over a particle which displaces it from its origin to the point $\vec{r}= \left ( 2\hat{i} -\hat{j}\right )m$ The work done on the particle in joule is

-7

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Concepts Covered - 2

Work done by a constant force

Work-

Work is said to be done when a force applied on the body displaces the body through a certain distance along the direction of the force.

Work done by a constant force-

The scalar product of the force vector ( $\underset{F}{\rightarrow}$ ) and the displacement vector ( $\underset{S}{\rightarrow}$ )

$W=\vec{F}.\vec{S}$

The product of the magnitude of force $\left ( F \right )$ magnitude of displacement $\left (S \right )$ and cosine of the angle between them $\left (\Theta \right )$

$W=FS\cos \Theta$

If the number of forces $\vec{F}_{1},\vec{F}_{2},\vec{F}_{3} ......,\vec{F}_{n}$ , are acting on a body and it shifts from position vector $\vec{r_1}$ to position vector $\vec{r_2}$

Then $W=\left(\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3} \ldots \ldots \ldots+\vec{F}_{n}\right) \cdot\left(\vec{r}_{2}-\vec{r}_{1}\right)=\vec{F}_{n e t} \cdot \vec{r}_{n e t}$

Units-

SI Unit-Joule
CGS Unit- Erg
$1 \ Joule = 10^7 Erg$

Dimension- $ML^2T^{-2}$
Dependence of work done by a constant force

Frame of reference

With a change of frame of reference (inertial) force does not change while displacement may change. So the work done by a force will be different in different frames.

i.e. A person is pushing a box inside a moving train with a force $\vec{F}$

Displacement inside train $\vec{S}$

Displacement of the train in the ground frame is $\vec{S}_{0}$

Then work done by the force $\vec{F}$ is $W= \vec{F}\cdot \left ( \vec{S}+\vec{S_{0}} \right )$

Nature of Work Done

Positive Work-

Positive work means that force (or its component) is parallel to displacement.
Means $0\leq \Theta < \frac{\pi }{2}$

Where $\Theta$ is the angle between force vectors and displacement vector

Maximum work = $W_{max} = FS,\ When\ \theta = 0^{0}$
E.g When you move a block by pulling it then work done by you on the block is positive

Negative Work

Negative work means that force (or its component) is opposite to displacement.
Means $\frac{\pi }{2}< \Theta \leq \pi$

Where $\Theta$ is the angle between force vectors and displacement vector

Minimum work= $W_{min} = -FS,\ When\ \theta = 180^{0}$
E.g When a body is made to slide over a rough surface, the work done by the frictional force is negative