Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the c

NCERT Exemplar · Q25

Q. Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.

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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 …

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Work done by a force is calculated using the dot product of the force and displacement vectors. Since the force of gravity acts vertically downwards and the car's displacement is purely horizontal, the angle between them is $90^\circ$ , making the work done against gravity zero.

Work, in physics, is a precise concept. It's not just about effort; it's about a force causing a displacement in its own direction. When a force acts on an object, and the object moves, work is done. The key here is the direction of both the force and the displacement.

The work done ( $W$ ) by a constant force ( $\vec{F}$ ) causing a displacement ( $\vec{d}$ ) is defined as the dot product of these two vectors:

W = \vec{F} \cdot \vec{d}

This can also be expressed as:

W = |\vec{F}| |\vec{d}| \cos \theta

where $|\vec{F}|$ is the magnitude of the force, $|\vec{d}|$ is the magnitude of the displacement, and $\theta$ is the angle between the force vector and the displacement vector.

Important

Work is a scalar quantity, but it depends on the vector nature (direction) of both force and displacement. The angle $\theta$ is crucial.

In this problem, we are asked to calculate the work done against gravity. This means we are considering the work done by an external agent (the car's engine, indirectly) to overcome the force of gravity. However, if the force of gravity itself does no work, then no work needs to be done against it either. Let's analyze the work done by gravity first.

Here's how we approach this problem:

Identify the force of gravity:

The force of gravity ( $\vec{F}_g$ ) acts on the car. Its magnitude is $F_g = mg$ , where $m$ is the mass of the car and $g$ is the acceleration due to gravity.
- $m = 400 \text{ kg}$
- $g \approx 9.8 \text{ m/s}^2$ (or $10 \text{ m/s}^2$ for simpler calculations, but it won't matter here). The direction of the force of gravity is always vertically downwards.
Identify the displacement:

The car moves along a straight horizontal road.
- The magnitude of the displacement is $d = 2 \text{ m}$ . The direction of the displacement is horizontally.
Determine the angle between the force and displacement:

The force of gravity acts vertically downwards. The displacement is purely horizontal. A vertical direction is always perpendicular to a horizontal direction. …

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Q.Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.

Concept understanding — Dot Product Angle

The Dot Product and the Angle Between Vectors

Step 1: What is a dot product?

Step 2: The geometric meaning — the angle connection

Step 3: Why does this make sense?

Step 4: Finding the angle from the dot product

Step 5: The precise statement …

Concept: Work as a Dot Product — Perpendicular Force and Displacement

Q.Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.

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