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Work and Power
The Physics Definition of Work
Work is defined as the scalar product of force and displacement, measuring the energy transferred by a force acting over a distance.
$$W = \vec{F} \cdot \vec{s} = Fs \cos \theta$$
Work is the component of force in the direction of motion multiplied by displacement.
$W$=Work done(Joules (J))
$F$=Magnitude of applied force(Newtons (N))
$s$=Magnitude of displacement(Meters (m))
$\theta$=Angle between force and displacement vectors(degrees)
$\theta = 0^\circ$
→$W = Fs$ (Maximum positive work)
$\theta = 90^\circ$
→$W = 0$ (No work done)
Force Component: Only the component $F \cos \theta$ along the displacement vector contributes to work.
Unit Definition: One Joule is the work done by a 1 N force moving an object 1 m in the force's direction.
Nature of Work: Directional Dependency
The sign of work depends on the relative orientation of force and motion, determining if energy is added to or removed from a system.
$$W = 0 \text{ if } \theta = 90^\circ$$
Positive Work: Occurs when $0^\circ \le \theta < 90^\circ$ (force aids motion).
Negative Work: Occurs when $90^\circ < \theta \le 180^\circ$ (force opposes motion, e.g., friction).
Zero Work: Occurs when force is perpendicular to displacement (e.g., Centripetal Force or normal force on a horizontal plane).
Zero Work Scenarios
Zero work occurs when there is no displacement component in the direction of force.
$$W = 0 \text{ when } \vec{F} \perp \vec{s}$$
Zero work occurs when force is perpendicular to displacement, OR when displacement is zero.
$\theta = 90^\circ$=Force perpendicular to motion(degrees)
$s = 0$=No displacement occurs(m)
Satellite in circular orbit
→Gravity does $W = 0$ (centripetal force)
Pendulum bob swings
→Tension does $W = 0$ (always perpendicular to motion)
Car on horizontal road
→Normal force does $W = 0$ (perpendicular to motion)
Pushing a wall: Force applied but $s = 0$, so $W = 0$.
Circular motion: Centripetal Force is always perpendicular to tangential velocity.
Work Done by a Variable Force
When force changes with position, total work is calculated as the integral of the force function over the displacement path.
$$W = \int_{x_1}^{x_2} F(x) dx$$
Work equals the area under the curve of a Force vs. Displacement graph.
$F(x)$=Force as a function of position(N)
$dx$=Infinitesimal displacement element(m)
$x_1, x_2$=Initial and final positions(m)
$F = kx$
→$W = \frac{1}{2}kx^2$ (Work done by a spring)
Graphical Method: Area above the x-axis represents positive work; area below represents negative work.
Summation Approach: Total work is the limit of the sum of $F_i \cos \theta_i \Delta d_i$ as $\Delta d$ approaches zero.
Power: The Rate of Energy Transfer
Power quantifies how quickly work is performed or energy is converted.
$$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$$
Power is the time-derivative of work or the dot product of force and velocity.
$P$=Instantaneous power(Watts (W))
$dW$=Small amount of work done(J)
$dt$=Time interval(s)
$v$=Instantaneous velocity(m/s)
Constant $v$
→$P = Fv$ (Power required to overcome resistance)
Standard Units: The Watt (1 J/s). Larger units include horsepower (1 hp ≈ 746 W).
Average vs. Instantaneous: Average power uses total work/time; instantaneous power is the rate at a specific moment.
The Work-Energy Theorem
The Work-Energy Theorem connects force-based mechanics to energy-based analysis. It states that net work equals change in Kinetic Energy.
$$W_{net} = \Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
The work done by all forces (net work) equals the change in kinetic energy.
$W_{net}$=Sum of work done by all forces(J)
$\Delta K$=Change in kinetic energy(J)
$K = \frac{1}{2}mv^2$=Kinetic energy(J)
Positive net work
→$KE$ increases ($v_f > v_i$)
Negative net work
→$KE$ decreases ($v_f < v_i$)
Zero net work
→$KE$ unchanged (speed constant)
Sign Rule: If an object speeds up, $W_{net} > 0$. If it slows down, $W_{net} < 0$.
Friction Work: Always negative (reduces $KE$).
Gravity Work: Depends only on vertical height change: $W_g = mgh$, regardless of the path taken.
Dimensional Analysis
Dimensional analysis provides a way to verify equations and check answer consistency.
$$[W] = [ML^2T^{-2}], \quad [P] = [ML^2T^{-3}]$$
Work and power dimensions derived from their defining formulas.
$[W] = [F][s] = [MLT^{-2}][L]$=Work dimension from force × displacement($[ML^2T^{-2}]$)
$[P] = [W]/[t]$=Power dimension from work/time($[ML^2T^{-3}]$)
Checking formula validity
→Both sides must have same dimensions
Converting units
→1 J = $10^7$ erg (CGS to SI)
Quick Check: Work and Energy have same dimensions: $[ML^2T^{-2}]$.
Power vs. Work: Power has one extra $T^{-1}$ (time in denominator).
Proportionality Relationships
Proportionality analysis lets you determine how changes in one variable affect another without full calculations.
$$W \propto F, \quad W \propto s, \quad W \propto \cos \theta$$
From $W = Fs \cos \theta$: If $F$ doubles and $s, \theta$ constant, $W$ doubles.
From $P = W/t$: If same work in half time, $P$ doubles.
From $P = Fv$: If $v$ triples and $F$ constant, $P$ triples.
Application Strategy
Worked Examples
Problem Statement
A box is pulled 5 meters across a floor by a rope applying a 20 N force at an angle of 30° to the floor. Find the work done.
Work formula $W = Fs \cos \theta$
Strategic Toolkit
$W = Fs \cos \theta$
Step 1
Identify Givens
Extract force, displacement, and angle from the problem statement.
$F = 20 \text{ N}, s = 5 \text{ m}, \theta = 30^\circ$
Step 2
Formula Setup
Substitute the values into the work definition equation.
$W = 20 \times 5 \times \cos(30^\circ)$
Step 3
Calculate Result
Solve for W using $\cos(30^\circ) \approx 0.866$.
$W = 100 \times 0.866 = 86.6 \text{ J}$
Problem Statement
A car engine provides a force of 2000 N to maintain a constant speed of 20 m/s. What is the power output of the engine?
Power-Velocity Relationship $P = Fv$
Strategic Toolkit
P = Fv
Step 1
Identify Givens
We have a constant force acting in the direction of velocity.
$F = 2000 \text{ N}, v = 20 \text{ m/s}$
Step 2
Calculate Power
Multiply force by velocity for instantaneous power.
$P = 2000 \times 20 = 40,000 \text{ W}$
Step 3
Unit Conversion
Express the result in kilowatts for better readability.
$P = 40 \text{ kW}$
Problem Statement
A 1000 kg car accelerates from 10 m/s to 20 m/s. How much net work was done on the car?
Work-Energy TheoremKinetic Energy
Strategic Toolkit
$W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
Step 1
Identify Givens
Extract mass and initial/final velocities.
$m = 1000 \text{ kg}, v_i = 10 \text{ m/s}, v_f = 20 \text{ m/s}$
Step 2
Calculate Initial KE
Find the initial kinetic energy.
$K_i = \frac{1}{2}(1000)(10)^2 = 50,000 \text{ J}$
Step 3
Calculate Final KE
Find the final kinetic energy.
$K_f = \frac{1}{2}(1000)(20)^2 = 200,000 \text{ J}$
Step 4
Apply Work-Energy Theorem
Net work equals change in kinetic energy.
$W_{net} = K_f - K_i = 200,000 - 50,000 = 150,000 \text{ J}$
Memorize formulas
Work Definition
Use when force is constant and acting at an angle theta to the movement.
Formula
$W = Fs \cos \theta$
Work Vector Form
Use when Force and Displacement are given in vector (i, j, k) notation.
Formula
$W = \vec{F} \cdot \vec{s}$
Power (General)
The rate at which work is done over time.
Formula
$P = \frac{W}{t}$
Power (Kinetic)
Mnemonic: 'Fast Vehicles' - use when you have constant force and constant velocity.
Formula
P = Fv
Variable Work
Use when the force changes depending on the position x.
Formula
$W = \int F(x) dx$
Work-Energy Theorem
Net work equals change in kinetic energy.
Formula
$W_{net} = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
Spring Work
Work done to stretch/compress a spring by distance x.
Formula
$W = \frac{1}{2}kx^2$
Gravity Work (Incline)
NUST shortcut: Work against gravity depends only on height change!
Formula
$W_g = mgh$
Review
1
A satellite moves in a circular orbit around the Earth. What is the work done by the gravitational force on the satellite?
Ans
Zero, because the force is always perpendicular to the instantaneous displacement
Review
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Use the perpendicular-force rule
Gravity provides the centripetal force, while the satellite’s instantaneous displacement is tangential to the orbit. The angle between force and displacement is therefore $90^\circ$, so the force does no work.
$$W = F s \cos 90^\circ = 0$$
2
A man pushes a wall with a force of 500 N for 10 minutes but the wall does not move. How much work is done?
Ans
Zero
Review
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Check whether any displacement occurs
Work is only done when a force produces displacement. Since the wall does not move, the displacement is zero even though the applied force and time are large.
$$W = Fs = 500 \times 0 = 0$$
3
A car of mass 1200 kg is moving at a constant velocity of 15 m/s up a hill inclined at 30° to the horizontal. If friction is negligible, what power is the engine delivering? (Use $g = 10$ m/s²)
Ans
90,000 W
Review
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Find the force needed up the slope
Because the car moves at constant velocity and friction is negligible, the engine only needs to balance the component of weight along the incline.
$$F = mg\sin 30^\circ = 1200 \times 10 \times 0.5 = 6000\ \text{N}$$
Use power as force times speed
When the driving force is along the direction of motion, power is the product of force and velocity.
$$P = Fv = 6000 \times 15 = 90{,}000\ \text{W}$$
4
A force of 50 N acts on an object, displacing it by 8 m. If the angle between the force and the displacement is 60°, the work done is:
Ans
200 J
Review
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Resolve the force along the displacement
Only the component of the force in the direction of motion contributes to work, so use the cosine of the angle between force and displacement.
$$W = Fd\cos\theta = 50 \times 8 \times \cos 60^\circ = 200\ \text{J}$$
5
If the power of a machine is constant, the velocity of a body of mass $m$ starting from rest varies with time $t$ as:
Ans
$v \propto \sqrt{t}$
Review
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Write constant power in terms of velocity
For a body of mass $m$, the driving force is $F = m\,dv/dt$, so constant power becomes $P = Fv = m(dv/dt)v$.
$$P = m\,v\frac{dv}{dt}$$
Integrate the separated variables
Separate variables and integrate from rest. This gives $v^2 \propto t$, so velocity grows with the square root of time.
$$v\,dv = \frac{P}{m}\,dt \;\Rightarrow\; \frac{v^2}{2} = \frac{P}{m}t \;\Rightarrow\; v \propto \sqrt{t}$$
6
A 100 kg box is pushed up a 30° incline for a distance of 10 m along the incline. The vertical height gained is 5 m. How much work is done against gravity? (Use $g = 10$ m/s²)
Ans
5000 J
Review
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Use the height gained, not the path length
The work done against gravity depends on the change in gravitational potential energy, which depends only on vertical height.
$$W_g = mgh = 100 \times 10 \times 5 = 5000\ \text{J}$$
7
A spring with $k = 500$ N/m is already stretched by 0.1 m from equilibrium. How much additional work is needed to stretch it further to 0.3 m from equilibrium?
Ans
20 J
Review
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Use the change in spring energy
Additional work is the increase in elastic potential energy from the initial stretch to the final stretch, not the energy from zero to the final position.
$$W = \frac{1}{2}k(x_2^2 - x_1^2)$$
Substitute both stretches
Here $x_1 = 0.1\,\text{m}$ and $x_2 = 0.3\,\text{m}$, so the square difference gives the net extra work needed.
$$W = \frac{1}{2}(500)(0.3^2 - 0.1^2) = 250(0.09 - 0.01) = 20\ \text{J}$$
8
If force is doubled and displacement is halved, the work done:
Ans
Remains the same
Review
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Apply the proportional changes directly
Work is the product of force and displacement. Doubling one factor while halving the other leaves the product unchanged.
$$W_{new} = (2F)\left(\frac{s}{2}\right) = Fs = W$$
9
A car engine produces 40 kW of power. If the velocity is doubled from 10 m/s to 20 m/s (assuming constant force), what happens to the power?
Ans
Power doubles to 80 kW
Review
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Use the force-velocity relation
With constant force, power is directly proportional to velocity, so any change in speed produces the same proportional change in power.
$$P = Fv \;\Rightarrow\; v \times 2 \implies P \times 2$$
10
1 horsepower (hp) is approximately equal to:
Ans
746 W
Review
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Recall the standard conversion
Horsepower is a standard power unit, and its commonly used mechanical-equivalent value is about 746 watts.
$$1\ \text{hp} \approx 746\ \text{W}$$
Question 1 of 10
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1
A satellite moves in a circular orbit around the Earth. What is the work done by the gravitational force on the satellite?