When does the work done by a force negative?

Question of 83

Q. When does the work done by a force negative?

Karnataka PUCKarnataka School Examination and Assessment Board (1st PUC) 2020Subjective· 1mImportance★★★★★

0% · 0/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Dot Product Angle

The Dot Product and the Angle Between Vectors

Imagine you're pushing a heavy box across the floor. You push at an angle — not straight forward, but partly downward and partly forward. The part of your push that actually moves the box is only the forward component. The downward part just presses the box into the floor.

That's the core intuition behind the dot product: it measures how much one vector "goes in the direction of" another vector.

Step 1: What is a dot product?

Given two vectors $\vec{a}$ and $\vec{b}$ in 2D or 3D space, their dot product (also called the scalar product) is defined algebraically as:

\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

You multiply corresponding components and add them up. The result is a single number (a scalar), not a vector.

For example, if $\vec{a} = (3, 4)$ and $\vec{b} = (2, -1)$ , then:

\vec{a} \cdot \vec{b} = 3 \times 2 + 4 \times (-1) = 6 - 4 = 2

Step 2: The geometric meaning — the angle connection

Here's the beautiful part. The dot product also has a completely different geometric definition:

\vec{a} \cdot \vec{b} = |\vec{a}| \, |\vec{b}| \, \cos \theta

where $|\vec{a}|$ and $|\vec{b}|$ are the magnitudes (lengths) of the vectors, and $\theta$ is the angle between them when they're placed tail-to-tail.

This is the dot product angle formula. It connects algebra (component multiplication) to geometry (angle and length).

Step 3: Why does this make sense?

Think about the extreme cases:

Vectors point in the same direction ( $\theta = 0^\circ$ ): $\cos 0 = 1$ , so $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}|$ — the maximum possible value. All of one vector's "push" is in the other's direction.
Vectors are perpendicular ( $\theta = 90^\circ$ ): $\cos 90^\circ = 0$ , so $\vec{a} \cdot \vec{b} = 0$ . Neither vector has any component along the other. This is a crucial test for orthogonality.
Vectors point opposite ( $\theta = 180^\circ$ ): $\cos 180^\circ = -1$ , so $\vec{a} \cdot \vec{b} = -|\vec{a}| |\vec{b}|$ — the most negative value. They're completely against each other.
Any other angle: the dot product is somewhere between these extremes, proportional to how much one vector "projects" onto the other.

Tip

The dot product is positive when the angle is acute ( $< 90^\circ$ ), zero when perpendicular, and negative when obtuse ( $> 90^\circ$ ). This sign alone tells you whether the vectors are generally aligned or opposed.

Step 4: Finding the angle from the dot product

If you know the components of two vectors, you can find the angle between them by rearranging the formula:

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \, |\vec{b}|}

Then use $\theta = \cos^{-1}(\text{that value})$ .

Example: Find the angle between $\vec{a} = (1, 2)$ and $\vec{b} = (3, 4)$ .

Compute dot product: $1 \times 3 + 2 \times 4 = 3 + 8 = 11$
Compute magnitudes: $|\vec{a}| = \sqrt{1^2 + 2^2} = \sqrt{5}$ , $|\vec{b}| = \sqrt{3^2 + 4^2} = 5$
$\cos \theta = \frac{11}{\sqrt{5} \times 5} = \frac{11}{5\sqrt{5}} \approx 0.9839$
$\theta = \cos^{-1}(0.9839) \approx 10.3^\circ$

The vectors are nearly aligned.

Watch out

The dot product formula gives $\cos \theta$ , not $\theta$ itself. Always take the inverse cosine. Also, the formula works for vectors of any dimension — 2D, 3D, even 100D — as long as you use the component definition.

Step 5: The precise statement …

Start your 14-day free trial to unlock the full solution →

[!TLDR]

Work done is negative when the angle θ between the force and the displacement is obtuse (90° < θ ≤ 180°), i.e. the force has a component opposite to the displacement.

Method …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. An object moves on a smooth inclined plane without slipping. The work done by the inclined plane surface on the ball is$
(a) Positive
(b) Negative
(c) Zero
(d) None of these
›Reveal solutionSolution
Work done = Force x displacement x cos(angle between them). The normal reaction from the incline is always perpendicular (90 degrees) to the direction of motion along the incline, so its work is zero.

Work done by a force is W = F d cos(theta), where theta is the angle between the force and the displacement.

For an object sliding/rolling without slipping on a SMOOTH inclined plane:

The normal force N from the surface always acts perpendicular to the inclined surface.

The object's displacement, as it moves along the incline, is always parallel to the inclined surface.

Therefore the angle between N and the displacement is always 90 degrees, and cos(90) = 0. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. A force F = 2i^ + 3j^ + k^ acts on a body. The work done by the force for a displacement of -2i^ + j^ - k^ is$
(a) 2 units
(b) 4 units
(c) -2 units
(d) -4 units
›Reveal solutionSolution

Work done by a constant force for a given displacement is the dot product W = F . d; carrying out the component-wise multiplication and summing gives -2 units.

Given:

F = 2i + 3j + k

d = -2i + j - k

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. If a Force of 10 N moves an object 15m in the direction of Force. How much work is done.$
(a) 1.5 J
(b) 150 J
(c) 100 J
(d) 0.66 J.
›Reveal solutionSolution

Since the force and displacement are in the same direction, W = F × d = 10 N × 15 m = 150 J.

Work done by a force is defined as:

W = F·d·cosθ

Here the force moves the object in its own direction, so θ = 0° and cosθ = 1.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. When force acts along the direction of displacement then the work done is ___________.$
›Reveal solutionSolution
[!TLDR]

maximum

Method

W = Fd cosθ; θ = 0° gives cosθ = 1, the maximum possible value, so …
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. A body constrained to move along y-axis is subjected to a constant force F = (-i + 2j + 3k) N. The work done by this force in moving the body a distance of 4 m along y-axis is$
(1) 4 J
(2) 8 J
(3) 12 J
(4) 24 J
›Reveal solutionSolution

Work done = F . d; since the displacement is purely along y, only the y-component of the force contributes.

Given: F = (-i + 2j + 3k) N, and the body moves a distance d = 4 m purely along the y-axis, so displacement vector s = 4j m.

Work done:

W = F . s = (-i + 2j + 3k) . (4j) = (-1)(0) + (2)(4) + (3)(0) = 8 J

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Product of force and displacement is:$
(a) vector
(b) scaler
(c) none of the two
(d) only number
›Reveal solutionSolution

Work = F·d is a scalar.

Work is defined as the scalar (dot) product of the force and displacement vectors: W = F·d = Fd cosθ. Although F and d are vectors, their dot product yields a scalar quan …

Start your 14-day free trial to see the full solution →
CBSE 2022Set TERM11 markMCQ
$Q. A body is being raised to a height h from the surface of earth. What is the sign of work done by$
(i) applied force and
(ii) gravitational force respectively?
(1) Positive, Positive
(2) Positive, Negative
(3) Negative, Positive
(4) Negative, Negative
›Reveal solutionSolution

Work done, W = F.d.cos(theta), is positive when force and displacement point the same way, negative when they point opposite ways. Here displacement is upward: the applied force (upward) matches it (positive work); gravity (downward) opposes it (negative work).

The body is raised through a height h, so its displacement is directed upward.

(i) Applied force: to lift the body, the applied force must act upward, in the SAME direction as the displacement. The angle between them is 0°, so cos(0°) = +1, giving positive work: W_applied = F.h > 0.

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set TERM11 markMCQ
$Q. The work performed on an object does not depend upon$
(1) the displacement
(2) the force applied
(3) the angle between force and displacement
(4) initial velocity of the object
›Reveal solutionSolution

By its very definition, W = F.d.cos(theta), work depends only on the applied force, the displacement it causes, and the angle between them -- the object's initial velocity plays no role in this formula.

Work done by a force is defined as:

W = F . d . cos(theta)

where F is the magnitude of the force, d is the magnitude of the displacement, and theta is the angle between the force and displacement vectors.

This definition explicitly involves:

(1) the displacement -- YES, it appears directly.

(2) the force applied -- YES, it appears directly. …

Start your 14-day free trial to see the full solution →
CBSE 2022Set TERM11 markMCQ
$Q. Work is represented by$
(1) W = F x S
(2) W = S x F
(3) W = F . S
(4) none of these
›Reveal solutionSolution

Work must come out as a scalar even though both force and displacement are vectors -- this is achieved precisely by the dot product, W = F.S, not by a cross product or an ordinary (undefined) product of vectors.

Force (F) and displacement (S) are both vectors, but work (W) is a SCALAR quantity (it has magnitude only, no direction). The only well-defined way to combine two vectors into a scalar is the dot (scalar) product:

W = F . S = |F| |S| cos(theta)

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set TERM11 markMCQ
$Q. A force does maximum work on an object. The angle between the force and displacement vector is:$
(a) 0°
(b) 30°
(c) 60°
(d) 90°
›Reveal solutionSolution

W = Fd cos(theta) is maximum when cos(theta) is maximum (cos(theta) = 1), which happens at theta = 0 degrees, i.e. when force and displacement are in the same direction.

Work done by a constant force F causing a displacement d, with the angle between F and d being theta, is:

W = F d cos(theta)

…

Start your 14-day free trial to see the full solution →
CBSE 2021Set TERM11 markMCQ
$Q. A Loader carries 25 kg load on his head and travels 200m on a straight horizontal road. The work done by the Loader is:$
(a) Infinite
(b) Positive, but not infinite
(c) Negative
(d) None of these
›Reveal solutionSolution

The loader's force on the load (upward, to support its weight) is at 90 degrees to the horizontal displacement; since W = Fd cos(theta) and cos(90 degrees) = 0, the work done is exactly zero -- not listed among (a) infinite, (b) positive-finite, or (c) negative, so the correct choice is (d) None of these.

The loader supports the 25 kg load by exerting an upward force (equal in magnitude to the load's weight, mg) on his head. He then walks 200 m along a straight, horizontal road.

Work done, W = F d cos(theta), where theta is the angle between the applied force and the displacement.

Here the applied force is vertical (upward) and the displacement is horizontal, so theta = 90 degrees, and cos(90 degrees) = 0. Therefore:

…

Start your 14-day free trial to see the full solution →
CBSE 2020Set ANNUAL1 markMCQ
$Q. Work is represented by$
(a) W = F x S (both vectors)
(b) W = S x F (both vectors)
(c) W = F . S (both vectors)
(d) none of these
›Reveal solutionSolution

Work done is defined as the dot product of the force vector and the displacement vector, which yields a scalar quantity.

Work is defined as W = F . S = |F||S| cos(theta), where theta is the angle between the force and displacement vectors. The dot product ensures only the component of force along the displacement contributes to work, and the result is a scalar (work has no direction).

…

Start your 14-day free trial to see the full solution →

←Previous chapterCh 4Laws of Motion

Q.When does the work done by a force negative?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days