JEE Mains Physics · Work, Energy and Power

Work Done by Constant and Variable Forces

Work is force times displacement along the force: a dot product for a constant force, and an integral of F dx when the force changes with position.

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Why this matters

Nineteen PYQs, nine of them multiple choice, and one from 2026. Five take the dot product of a constant force with a displacement or judge the sign of a work; ten integrate a force that changes with position, six of them along a line, two in a plane, one as areas under F–x graphs and one with a force that turns as it moves; four slow a block with a force that grows with distance. More than half ask for a number, so the arithmetic has to be exact.

Concept 1 of 3: Work done by a constant force

Only the part of a force that lies along the displacement does work. That is why work is a dot product: a force along the motion does positive work, a force against it does negative work, and a force at right angles does none. Work is a scalar, so with several forces you may add the forces first or add their works; the answer is the same.

Definition

$W = \vec F\cdot\vec s = Fs\cos\theta = F_x s_x + F_y s_y + F_z s_z$ .
The displacement is final position minus initial position: $\vec s = \vec r_2 - \vec r_1$ .
A distance d along a direction $\vec a$ is $\vec s = d\,\dfrac{\vec a}{|\vec a|}$ : 10 m along $3\hat i + 4\hat j$ is $6\hat i + 8\hat j$ .
Several forces: $W = (\vec F_1 + \vec F_2)\cdot\vec s = W_1 + W_2$ .
Sign: $\theta < 90^{\circ}$ positive, $\theta = 90^{\circ}$ zero, $\theta > 90^{\circ}$ negative.
Gravity on a bucket being lifted does negative work; the person lifting does positive work. Kinetic friction on a sliding body and air resistance on a swinging pendulum do negative work.
Zero work means $\vec F \perp \vec s$ : set $\vec F\cdot\vec s = 0$ to find an unknown component.

Work by a constant force

W = \vec F\cdot\vec s = Fs\cos\theta = F_x s_x + F_y s_y + F_z s_z

Worked example

A constant force

\vec F = (4\hat i - 2\hat j + 3\hat k)

N moves a particle from the point

(1, 2, 0)

m to the point

(3, -1, 2)

m. Find the work done.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 1 · Q21Moderate

Example 1 · Work, Energy and Power · Work Done by Constant and Variable Forces

A 1 kg block subjected to two simultaneous forces

(2\widehat{i} + 3\widehat{j} + 4\widehat{k})N

and

(3\widehat{i} - \widehat{j} - 2\widehat{k})N

is moved a distance of 25 m along

(3\widehat{i} - 4\widehat{j})

direction. The work done in this process is

\_\_\_\_

A distance along a direction needs the unit vector

Moving 10 m along 3î + 4ĵ means a displacement of 6î + 8ĵ, because |3î + 4ĵ| = 5. Multiplying the direction vector by 10 gives 30î + 40ĵ, a displacement five times too long.

A force that balances friction does positive work

A box pulled at constant velocity on a rough floor has zero net work done on it. The pull still does positive work; friction does an equal negative work. Saying the applied force does zero work confuses one force's work with the net work.

Drill 4 more on work done by a constant force

Concept 2 of 3: Work done by a force that varies with position

When the force changes as the body moves, cut the path into small steps. Each step does F dx of work, and adding the steps is an integral. On an F–x graph the same sum is the area under the curve. In a plane, the x-part of the force works only through dx and the y-part only through dy.

Definition

One dimension: $W = \displaystyle\int_{x_1}^{x_2} F(x)\,dx$ .
On an F–x graph, W is the area between the curve and the x-axis. Area below the axis counts as negative work.
In a plane: $W = \displaystyle\int \vec F\cdot d\vec r = \int F_x\,dx + \int F_y\,dy$ .
If $F_x$ depends only on x and $F_y$ only on y, each integral runs between its own end values and the path does not matter.
If $F_x$ contains y (such as $x^{2}y$ ), the work depends on the path: put the path's equation in before integrating.
A force of fixed size F whose direction makes an angle $\theta(x)$ with the motion: $W = \displaystyle\int F\cos\theta(x)\,dx$ .
When this is the only force that does work, the same integral is the change in kinetic energy.

Work by a variable force

W = \int_{x_1}^{x_2} F(x)\,dx \qquad W = \int F_x\,dx + \int F_y\,dy

Worked example

A force

F = (6x^{2} - 4x)

N acts along the x-axis. Find the work it does as the body moves from

x = 1

m to

x = 3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q19Moderate

Example 2 · Work, Energy and Power · Work Done by Constant and Variable Forces

A force

F = \alpha + \beta x^{2}

acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant,

\alpha = 1\text{ }N

then

\beta

will be

Area below the axis is negative work

On an F–x graph, the parts of the curve below the x-axis are a force opposing the motion. Their area is subtracted. Adding every area as positive puts the graphs in the wrong order.

A force like x²y î depends on the path

If the x-component of a force contains y, the integral of Fₓ dx cannot be done until y is written in terms of x along the path. Integrating it with y held constant gives a number that belongs to no path.

Drill 9 more on work done by a force that varies with position

Concept 3 of 3: Slowing down under a retarding force that depends on position

A rough patch or a damping force that grows with distance takes kinetic energy away, and the amount it takes is the work it does. So the speed after the patch comes from one energy balance, and the stopping point is where that work has used up all the starting kinetic energy.

Definition

Retarding force $F = -kx$ between $x = a$ and $x = b$ : work $= -\dfrac{k}{2}(b^{2} - a^{2})$ .
Speed after the patch: $\tfrac12 mv^{2} = \tfrac12 mu^{2} - \dfrac{k}{2}(b^{2} - a^{2})$ .
Stopping distance from the origin with $ma = -\alpha x^{n}$ : $\dfrac{\alpha x^{n+1}}{n+1} = \tfrac12 mv_0^{2}$ .
A retardation $a(x)$ given per unit mass: loss of kinetic energy $= m\displaystyle\int a\,dx$ .
The same result comes from $v\,\dfrac{dv}{dx} = a$ , integrated on both sides.

Energy after a position-dependent retarding force

\tfrac12 mv^{2} = \tfrac12 mu^{2} - \int_{a}^{b}|F(x)|\,dx

Worked example

A 2 kg block moving at 5 m/s enters a rough patch that runs from

x = 1

m to

x = 3

m. In the patch the retarding force is

F = -3x

N. Find the block's speed as it leaves the patch.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q94Moderate

Example 3 · Work, Energy and Power · Work Done by Constant and Variable Forces

A block of mass 1 kg , moving along

x

with speed

v_{i}= 10\text{ }m/s

enters a rough region ranging from

x = 0.1\text{ }m

x = 1.9\text{ }m

. The retarding force acting on the block in this range is

F_{r}= - kxN

, with

k = 10\text{ }N/m

. Then the final speed of the block as it crosses rough region is.

Use the patch's end points, not its length

For F = −kx between x = 1 m and x = 3 m, the work is (k/2)(3² − 1²) = 4k. Using the length, (k/2)(2)² = 2k, treats the patch as if it began at the origin.

A retarding force takes energy away

Its work is subtracted from the starting kinetic energy. Adding it gives a final speed larger than the starting speed, which a slowing force can never produce.

Drill 3 more on slowing down under a retarding force that depends on position

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Work done by a constant force
Work by a constant force
$W = \vec F\cdot\vec s = Fs\cos\theta = F_x s_x + F_y s_y + F_z s_z$
Work done by a force that varies with position
Work by a variable force
$W = \int_{x_1}^{x_2} F(x)\,dx \qquad W = \int F_x\,dx + \int F_y\,dy$
Slowing down under a retarding force that depends on position
Energy after a position-dependent retarding force
$\tfrac12 mv^{2} = \tfrac12 mu^{2} - \int_{a}^{b}|F(x)|\,dx$

Watch out for (6)

A distance along a direction needs the unit vector
→ Work done by a constant force
A force that balances friction does positive work
→ Work done by a constant force
Area below the axis is negative work
→ Work done by a force that varies with position
A force like x²y î depends on the path
→ Work done by a force that varies with position
Use the patch's end points, not its length
→ Slowing down under a retarding force that depends on position
A retarding force takes energy away
→ Slowing down under a retarding force that depends on position

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