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Work Done By A Constant Force - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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