Is work a scalar or a vector?

Work is a scalar. Although it is built from two vectors — force and displacement — it is defined as their scalar (dot) product, W = F·d = Fd cosθ. The product of two magnitudes and a cosine is a number, so work has magnitude and sign but no direction. NCERT states it plainly: though both force and displacement are vectors, work is a scalar.

Why is the work done in carrying a load horizontally zero?

When you carry a load at steady height across a room, the supporting force is vertical (upward) while the displacement is horizontal. The angle between them is 90°, and cos 90° = 0, so W = Fd cos 90° = 0. The force does no work because it has no component along the displacement, even though you feel tired — that tiredness is internal physiological energy, not physics work.

When is work positive, zero, and negative?

Work is positive when the angle θ between force and displacement lies between 0° and 90° (cosθ > 0), as when an applied force speeds up a body in its direction of motion. Work is zero when θ = 90° (cos 90° = 0) or when either the force or the displacement is zero. Work is negative when θ lies between 90° and 180° (cosθ < 0), as with friction or a braking force that opposes the motion (θ = 180°, cos 180° = −1).

What is the SI unit of work, and what is the erg?

The SI unit of work is the joule (J): one joule is the work done by a force of one newton acting through a displacement of one metre in the direction of the force, so 1 J = 1 N·m. The erg is the CGS unit of work, equal to one dyne acting through one centimetre; 1 erg = 10⁻⁷ J. The dimensional formula of work is [ML²T⁻²], the same as energy.

Does gravity do positive or negative work on a falling body?

When a body of mass m falls a height h, the gravitational force mg (downward) and the displacement (downward) are in the same direction (θ = 0°), so gravity does positive work W = +mgh. When the body rises through h, the displacement is upward while gravity still acts downward (θ = 180°), so gravity does negative work W = −mgh.

If a force is large, does it always do large work?

No. Work depends on displacement, not on force alone. NCERT notes that when you push hard against a rigid brick wall, the wall does not move, so the displacement is zero and the work done is zero however large the force. Likewise a weightlifter holding a 150 kg mass steadily on the shoulder for 30 s does no work on the load, because it does not move.

Can the work done by a force on a body equal the work done by the reaction on the other body?

Not necessarily. Newton's third law makes the forces equal and opposite, but work also depends on each body's displacement. In NCERT Example 5.3, a skidding cycle has the road do −2000 J on it, while the equal-and-opposite force the cycle exerts on the road does zero work because the road does not move. Equal-and-opposite forces do not imply equal-and-opposite work.

Work — Definition & Calculation — NEET Physics

Q: What is the SI unit of work, and what is the erg?

The SI unit of work is the joule (J): one joule is the work done by a force of one newton acting through a displacement of one metre in the direction of the force, so 1 J = 1 N·m. The erg is the CGS unit of work, equal to one dyne acting through one centimetre; 1 erg = 10⁻⁷ J. The dimensional formula of work is [ML²T⁻²], the same as energy.

Q: Does gravity do positive or negative work on a falling body?

When a body of mass m falls a height h, the gravitational force mg (downward) and the displacement (downward) are in the same direction (θ = 0°), so gravity does positive work W = +mgh. When the body rises through h, the displacement is upward while gravity still acts downward (θ = 180°), so gravity does negative work W = −mgh.

Q: If a force is large, does it always do large work?

No. Work depends on displacement, not on force alone. NCERT notes that when you push hard against a rigid brick wall, the wall does not move, so the displacement is zero and the work done is zero however large the force. Likewise a weightlifter holding a 150 kg mass steadily on the shoulder for 30 s does no work on the load, because it does not move.

Q: Can the work done by a force on a body equal the work done by the reaction on the other body?

Not necessarily. Newton's third law makes the forces equal and opposite, but work also depends on each body's displacement. In NCERT Example 5.3, a skidding cycle has the road do −2000 J on it, while the equal-and-opposite force the cycle exerts on the road does zero work because the road does not move. Equal-and-opposite forces do not imply equal-and-opposite work.

Everyday work vs physics work

NCERT opens the chapter by warning that the physical meaning of work is far narrower than the everyday one. A person who pushes hard against a rigid brick wall exerts a large force, tires, and burns internal energy — yet does no work on the wall, because the wall does not move. In physics, work requires both a force and a displacement of the point at which the force acts. Effort without displacement is not work.

NIOS makes the same point with the wall-pushing activity: irrespective of how hard you and your friends push, if the wall does not move, no work is done. The fatigue you feel is a physiological cost, not a measure of physics work. This sharp definition is the price of admission to the whole chapter — kinetic energy, potential energy and power all rest on it.

W = F · d = Fd cosθ

Work is the scalar product of force and displacement. The single number it produces carries a sign but no direction.

The scalar (dot) product

Force and displacement are both vectors, yet work must be a single number. The bridge is the scalar product. NCERT defines the scalar (dot) product of two vectors $\vec A$ and $\vec B$ as

$$\vec A \cdot \vec B = AB\cos\theta$$

where $\theta$ is the angle between them. Because $A$, $B$ and $\cos\theta$ are all scalars, the dot product is a scalar quantity — each vector has a direction, but their scalar product does not. Two useful readings follow at once: $\vec A \cdot \vec B = A(B\cos\theta) = B(A\cos\theta)$. Geometrically $B\cos\theta$ is the projection of $\vec B$ onto $\vec A$, so the dot product is the magnitude of one vector times the component of the other along it.

The product is commutative, $\vec A \cdot \vec B = \vec B \cdot \vec A$, and distributive over addition. In component form, for $\vec A = A_x\hat i + A_y\hat j + A_z\hat k$ and $\vec B = B_x\hat i + B_y\hat j + B_z\hat k$,

$$\vec A \cdot \vec B = A_xB_x + A_yB_y + A_zB_z$$

Two consequences anchor every work calculation: $\vec A \cdot \vec A = A^2$ (since $\theta = 0$), and $\vec A \cdot \vec B = 0$ when the vectors are perpendicular (since $\cos 90^\circ = 0$).

The scalar product $\vec A\cdot\vec B = AB\cos\theta$ equals the magnitude of $\vec A$ times the projection $B\cos\theta$ of $\vec B$ onto $\vec A$.

The definition of work

Consider a constant force $\vec F$ acting on an object that undergoes a displacement $\vec d$. Work done by the force is defined as the product of the component of force along the displacement and the magnitude of the displacement. With $\theta$ the angle between $\vec F$ and $\vec d$,

$$W = (F\cos\theta)\,d = \vec F \cdot \vec d$$

This is exactly the dot product of force and displacement. The component $F\cos\theta$ is the part of the force that points along the motion; the perpendicular component $F\sin\theta$ contributes nothing. NCERT motivates the same definition from kinematics: starting from $v^2 - u^2 = 2\,\vec a\cdot\vec d$ and multiplying by $m/2$ gives $\tfrac12 mv^2 - \tfrac12 mu^2 = m\,\vec a\cdot\vec d = \vec F\cdot\vec d$. The right-hand side, the product of displacement and the force-component along it, is named work.

Because work is defined as a dot product, it is a scalar. NCERT and NIOS both stress this: although force and displacement are vectors, work is a scalar. It can be positive, zero or negative, but it has no direction.

Geometry of W = Fd cosθ

Everything about the sign and size of work is contained in the angle $\theta$ between force and displacement, through the factor $\cos\theta$. Reading the cosine across its range gives the complete behaviour of work for a constant force.

Only the component $F\cos\theta$ along the displacement does work: $W = (F\cos\theta)\,d$. The perpendicular component $F\sin\theta$ does no work.

Angle θ	cosθ	Work W = Fd cosθ	Physical case
0°	+1	$+Fd$ (maximum positive)	Force fully along motion: accelerating a car forward
0° < θ < 90°	positive	positive	Force partly along motion: pulling a sledge by an inclined rope
90°	0	zero	Force perpendicular to motion: carrying a load horizontally
90° < θ < 180°	negative	negative	Force partly against motion
180°	−1	$-Fd$ (maximum negative)	Force fully against motion: friction or braking

Positive, zero and negative work

The factor $\cos\theta$ splits work into three regimes. Work is positive when $\theta$ lies between $0^\circ$ and $90^\circ$, because $\cos\theta > 0$. NIOS gives the cleanest case: a car driven forward by a force in its direction of motion ($\theta = 0^\circ$) speeds up, and $W = Fd\cos 0^\circ = +Fd$. A force that adds energy to a body does positive work.

Work is negative when $\theta$ lies between $90^\circ$ and $180^\circ$. When the brakes are applied, the force opposes the motion ($\theta = 180^\circ$) and $W = Fd\cos 180^\circ = -Fd$; the car loses speed. Friction opposing displacement is the standard negative-work case, with $\cos 180^\circ = -1$. A force that removes energy from a body does negative work.

Work is zero in three distinct situations, all worth memorising because NEET tests each. The displacement is zero (pushing the rigid wall). The force is zero (a block sliding on a frictionless table feels no horizontal force yet moves). Or the force and displacement are mutually perpendicular, $\theta = 90^\circ$, so $\cos 90^\circ = 0$ — as when you carry a load horizontally, where the supporting force is vertical, or when the Moon orbits the Earth in a circle and the radial gravitational pull is everywhere perpendicular to the tangential displacement.

Carrying a load at steady height: the support force is vertical, the displacement is horizontal, $\theta = 90^\circ$, so $W = Fd\cos 90^\circ = 0$.

Work done by gravity

Gravity is the cleanest laboratory for the sign of work, because the gravitational force $mg$ always points vertically down. NIOS works both cases. When a mass $m$ is lifted through a height $h$, the displacement is upward while gravity acts downward, so $\theta = 180^\circ$ and the work done by gravity is

$$W_{\text{grav}} = mg\,h\cos 180^\circ = -mgh$$

When the same mass is lowered through $h$, the force $mg$ and the displacement are both downward, $\theta = 0^\circ$, and

$$W_{\text{grav}} = mg\,h\cos 0^\circ = +mgh$$

So gravity does negative work on a rising body and positive work on a falling body. NIOS adds the careful counterpoint: when a person lifts the object, the work done by the person is $+mgh$ while the work done by gravity is $-mgh$; when lowering, the signs swap. The two are about different forces, both assuming the object moves without acceleration.

Work done by gravity is $-mgh$ on a rising body (force opposes displacement) and $+mgh$ on a falling body (force along displacement).

Newton's third law adds a final subtlety. In NCERT Example 5.3, the road exerts a stopping force of 200 N on a skidding cycle over 10 m, doing $W = 200\times10\times\cos 180^\circ = -2000\ \text{J}$ on the cycle. The cycle exerts an equal and opposite 200 N on the road, but the road does not move, so the work the cycle does on the road is zero. Equal-and-opposite forces do not imply equal-and-opposite work — work also depends on the displacement of each body.

Units of work

From $W = \vec F\cdot\vec d$, the unit of work is (unit of force) × (unit of displacement) = newton·metre. This is given the special name joule (J), after James Prescott Joule. One joule is the work done by a force of one newton acting through a displacement of one metre in the direction of the force:

$$1\ \text{J} = 1\ \text{N}\cdot\text{m}$$

The CGS unit is the erg — one dyne acting through one centimetre — listed by NCERT among the alternative units of work and energy, with $1\ \text{erg} = 10^{-7}\ \text{J}$. For practical electrical measurement NIOS gives the kilowatt-hour, $1\ \text{kW h} = 3.6\times10^{6}\ \text{J}$. Whatever the unit, the dimensional formula of work is $[\text{ML}^2\text{T}^{-2}]$, obtained from force $\times$ distance $= [\text{MLT}^{-2}]\times[\text{L}]$. Work and energy share these dimensions, which is why they share the joule.

Quantity	Symbol / unit	Relation
SI unit of work	joule (J)	`1 J = 1 N·m`
CGS unit of work	erg	`1 erg = 10⁻⁷ J`
Practical (electrical)	kilowatt-hour (kW h)	`1 kW h = 3.6 × 10⁶ J`
Dimensional formula	[ML²T⁻²]	force × distance

Worked numerical examples

Example 1 · NIOS 6.2

A force of 6 N is applied on an object at an angle of $60^\circ$ with the horizontal. Calculate the work done in moving the object 2 m in the horizontal direction.

Identify the angle. The force makes $60^\circ$ with the horizontal displacement, so $\theta = 60^\circ$ and $\cos 60^\circ = \tfrac12$.

Apply the definition. $W = Fd\cos\theta = 6\times 2\times\cos 60^\circ = 6\times 2\times\tfrac12 = 6\ \text{J}$.

Answer: the work done is $6\ \text{J}$. Only the horizontal component $F\cos 60^\circ = 3\ \text{N}$ does work over the 2 m displacement.

Example 2 · NIOS 6.3

A person lifts 5 kg of potatoes from the ground floor to a height of 4 m to bring them to the first floor. Calculate the work done. (Take $g = 9.8\ \text{m s}^{-2}$.)

Force needed. Lifting at constant speed, the applied force equals the weight: $F = mg = 5\times 9.8 = 49\ \text{N}$, directed up.

Work by the person. Force and displacement are both upward, $\theta = 0^\circ$: $W = Fd\cos 0^\circ = 49\times 4 = 196\ \text{J}$.

Answer: the person does $+196\ \text{J}$. Gravity, opposing the upward motion, does $-196\ \text{J}$ over the same lift.

Example 3 · NCERT Example 5.3

A cyclist comes to a skidding stop in 10 m. The force on the cycle due to the road is 200 N, directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?

(a) Road on cycle. The stopping (frictional) force and the displacement make $180^\circ$: $W = Fd\cos\theta = 200\times 10\times\cos 180^\circ = -2000\ \text{J}$. This negative work brings the cycle to a halt.

(b) Cycle on road. By Newton's third law the cycle pushes the road with 200 N, but the road undergoes no displacement, so the work done by the cycle on the road is $0$.

Answer: $-2000\ \text{J}$ and $0\ \text{J}$. Equal-and-opposite forces, but the work is not equal and opposite — displacement decides.

Work — Definition & Calculation