Work, Power, Kinetic Energy and Potential Energy
Work and the Scalar Product
Work is defined as the product of the magnitude of displacement and the component of the force in the direction of that displacement.
$$W = \mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$$
Determines the energy transferred to an object by a force acting over a distance.
$W$=Work done(Joules (J))
$F$=Magnitude of force(Newtons (N))
$d$=Displacement magnitude(meters (m))
$\theta$=Angle between Force and Displacement(degrees)
$\theta = 0°$
→$W = Fd$ (Maximum positive work — force along motion)
$\theta = 90°$
→$W = 0$ (No work done — force perpendicular to motion)
$\theta = 180°$
→$W = -Fd$ (Maximum negative work — force opposes motion)
Scalar Nature: Work is a scalar quantity even though it is derived from two vectors — the dot product always yields a number, not a direction.
Graphical Representation: The area under a Force-Displacement ($F \cos \theta$ vs $d$) graph equals the total work done.
Variable Forces: For changing forces, work equals the area under the $F\cos\theta$ vs $d$ curve: $W = \sum F_i \cos\theta_i \, \Delta d_i$.
Dimensional Check: $[W] = [F][d] = \text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{kg} \cdot \text{m}^2/\text{s}^2 = \text{J}$.
Work Done by Gravity and Conservative Forces
Work done by gravity on an object depends only on the vertical height change, not the path taken. This path-independence defines gravity as a conservative force.
$$W_{gravity} = -mgh$$
Work done by gravity when an object is raised by height $h$ (negative because gravity opposes upward displacement).
$m$=Mass of object(kg)
$g$=Gravitational acceleration(9.8 m/s²)
$h$=Vertical height change(m)
Object moves downward
→$W_{gravity} = +mgh$ (gravity does positive work)
Closed path (return to start)
→$W_{gravity} = 0$ (net height change is zero)
Path Independence: Whether an object goes straight up, along a ramp, or along a curved path — the work by gravity depends only on $\Delta h$.
Conservative Forces: Gravity, elastic spring force, and electric force are conservative — work done over any closed loop is zero.
Non-Conservative Forces: Friction, air resistance, tension, and applied propulsion forces are non-conservative — work depends on the path length.
Key Distinction: For conservative forces you can define a potential energy; for non-conservative forces you cannot.
Kinetic Energy and the Work-Energy Theorem
Kinetic energy is the energy of an object due to its motion, while the work-energy theorem states that net work done on a body equals the change in its $KE$.
$$W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$
Relates the total work performed by all forces to the resultant change in the object's speed.
$m$=Mass(kg)
$v_f$=Final velocity(m/s)
$v_i$=Initial velocity(m/s)
$W_{net} > 0$
→Object speeds up ($KE$ increases)
$W_{net} < 0$
→Object slows down ($KE$ decreases)
$W_{net} = 0$
→Speed is unchanged (constant velocity)
Velocity Dominance: $KE \propto v^2$ means doubling speed → 4× the kinetic energy, tripling speed → 9× the kinetic energy.
Net Work: Only the net (resultant) force contributes to the change in kinetic energy — individual forces may do positive or negative work.
Dimensional Check: $[KE] = \text{kg} \cdot (\text{m/s})^2 = \text{kg} \cdot \text{m}^2/\text{s}^2 = \text{J}$.
Potential Energy: Gravitational and Elastic
Gravitational potential energy near Earth's surface is energy stored due to an object's height above a chosen reference level.
$$U_g = mgh$$
Calculates energy stored due to vertical position in a uniform gravitational field.
$m$=Mass(kg)
$g$=Gravitational acceleration(9.8 m/s²)
$h$=Height above reference level(m)
$h = 0$
→Gravitational P.E. is zero (at reference level)
$h < 0$ (below reference)
→P.E. is negative — perfectly valid
Reference Level: P.E. is relative — only $\Delta U_g = mg\Delta h$ has physical meaning. Both students in the textbook are correct depending on their chosen reference.
Limitation: $U_g = mgh$ is valid only near Earth's surface where $g$ is approximately constant.
Elastic potential energy is energy stored in a spring (or any elastic object) due to its compression or extension from equilibrium.
$$U_s = \frac{1}{2}kx^2$$
Calculates energy stored in a spring displaced from its natural length.
$k$=Spring constant (stiffness)(N/m)
$x$=Displacement from equilibrium(m)
$x = 0$
→No stored energy (natural length)
Compression or extension
→Same $|x|$ stores the same energy — $x^2$ makes sign irrelevant
Restoring Force: Elastic P.E. comes from work done against the spring's restoring force $F = -kx$ (Hooke's Law).
Quadratic Dependence: $U_s \propto x^2$ means doubling the compression stores 4× the energy — not 2×.
Energy Exchange in SHM: In a mass-spring system, elastic PE and KE continuously interchange, but their sum $\frac{1}{2}kx_0^2$ is constant.
Conservation of Mechanical Energy
In an isolated system with only conservative forces, the total mechanical energy (sum of $KE$ and $PE$) remains constant throughout the motion.
$$KE_i + PE_i = KE_f + PE_f$$
Predicts the state of a system by assuming energy is transformed but never created or destroyed.
$KE$=Kinetic Energy(J)
$PE$=Potential Energy (Gravitational or Elastic)(J)
Non-conservative forces present
→$KE_i + PE_i = KE_f + PE_f + W_{friction}$
Object released from rest at height $h$
→Speed at bottom: $v = \sqrt{2gh}$ (mass cancels!)
Mass Cancels: For a falling object, $mgh = \frac{1}{2}mv^2$ → $v = \sqrt{2gh}$. The final speed is independent of mass — a feather and a bowling ball fall equally fast in vacuum.
With Friction: $mgh = \frac{1}{2}mv^2 + f \cdot d$ — friction steals energy and converts it to heat.
Percentage Lost: Energy lost to friction as a fraction: $\frac{f \cdot d}{mgh} \times 100\%$.
Interconversion: At any point during free fall: $PE + KE = mgh_{initial}$. As PE decreases, KE increases by exactly the same amount.
Power
Power is the rate of doing work, measuring how quickly energy is transferred or transformed.
$$P = \frac{W}{t} = Fv$$
Measures how fast energy is being transferred or transformed.
$P$=Power(Watts (W) = J/s)
$W$=Work done(Joules (J))
$t$=Time interval(seconds (s))
$v$=Velocity (constant)(m/s)
1 Horsepower
→≈ 746 Watts
1 kWh
→$= 3.6 \times 10^6$ J = 3.6 MJ (a unit of energy, not power)
Instantaneous Power: $P = \mathbf{F} \cdot \mathbf{v}$ gives power at any instant when force and velocity may vary.
Practical Application: Climbing stairs — same work ($mgh$) done quickly requires more power than doing it slowly.
Absolute Potential Energy and Escape Velocity
Far from Earth's surface, $g$ varies with distance. The absolute gravitational potential energy accounts for the $1/r$ dependence of the gravitational force.
$$U = -\frac{GMm}{r}$$
Gives the gravitational PE of mass $m$ at distance $r$ from Earth's center, with zero PE defined at infinity.
$G$=Universal gravitational constant($6.674 \times 10^{-11}$ N m²/kg²)
$M$=Mass of Earth($5.97 \times 10^{24}$ kg)
$m$=Mass of object(kg)
$r$=Distance from Earth's center(m)
$r = R$ (Earth's surface)
→$U_s = -\frac{GMm}{R}$ (most negative PE — deepest in gravitational well)
$r \to \infty$
→$U \to 0$ (object is free from gravitational influence)
Negative Sign: The negative sign means the object is gravitationally bound to Earth. Work must be done to increase $U$ toward zero.
Why Negative?: Zero PE is at infinity. Any finite $r$ means the object has fallen into the gravitational well, so PE is below zero.
Connection to $mgh$: Near the surface, $-\frac{GMm}{R+h} - (-\frac{GMm}{R}) \approx mgh$ when $h \ll R$.
The escape velocity is the minimum initial speed needed for an object to escape a planet's gravitational field entirely, reaching infinity with zero residual speed.
$$v_{esc} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$
Derived by equating initial KE to the absolute PE at the surface: $\frac{1}{2}mv_{esc}^2 = \frac{GMm}{R}$.
$v_{esc}$=Escape velocity(m/s)
$G$=Universal gravitational constant(N m²/kg²)
$M$=Mass of the planet(kg)
$R$=Radius of the planet(m)
Earth
→$v_{esc} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2$ km/s
Moon
→$v_{esc} \approx 2.4$ km/s (smaller mass and radius)
Mass Independence: $v_{esc}$ does not depend on the mass of the escaping object — a molecule and a rocket need the same escape speed.
Not Launch Speed: Real rockets don't need to reach 11.2 km/s instantaneously — they burn fuel continuously. Escape velocity applies to unpowered projectiles.
Derivation Shortcut: Set $KE = |U_{surface}|$: $\frac{1}{2}mv^2 = \frac{GMm}{R}$, the $m$ cancels immediately.
Application Strategy
Worked Examples
Problem Statement
A 1500 kg car traveling at 20 m/s brakes to a stop. How much work is done by the friction force to stop the car?
Work-Energy TheoremKinetic Energy
Strategic Toolkit
$$W_{net} = \Delta KE$$
Step 1
Identify Variables
Mass $m = 1500$ kg, initial velocity $v_i = 20$ m/s, final velocity $v_f = 0$ m/s.
$m=1500\text{ kg}, v_i=20\text{ m/s}, v_f=0\text{ m/s}$
Step 2
Calculate Initial Kinetic Energy
Use the $KE$ formula for the starting state.
$KE_i = \frac{1}{2}(1500)(20)^2 = 300{,}000\text{ J}$
Step 3
Apply Work-Energy Theorem
Work done is the change in kinetic energy ($KE_f - KE_i$). The negative sign confirms friction opposes motion.
$W = 0 - 300{,}000 = -300{,}000\text{ J} = -300\text{ kJ}$
Problem Statement
A spring with a constant of 400 N/m is compressed by 0.1 m. If it launches a 0.2 kg ball, what is the launch speed?
Elastic Potential EnergyConservation of Energy
Strategic Toolkit
$$PE_s = \frac{1}{2}kx^2, \quad KE = PE_s$$
Step 1
Calculate Stored Energy
Find the energy stored in the compressed spring.
$PE_s = \frac{1}{2}(400)(0.1)^2 = 2.0\text{ J}$
Step 2
Equate to Kinetic Energy
Assuming no losses, all elastic PE converts to KE.
$2.0 = \frac{1}{2}(0.2)v^2$
Step 3
Solve for Velocity
Rearrange and calculate the square root.
$v = \sqrt{\frac{2 \times 2.0}{0.2}} = \sqrt{20} \approx 4.47\text{ m/s}$
Problem Statement
A child starts from rest at the top of a 4.0 m high slide. If they reach the bottom at 6 m/s, what percentage of energy was lost to friction?
Conservation of EnergyNon-conservative Forces
Strategic Toolkit
$$mgh = \frac{1}{2}mv^2 + W_{friction}$$
Step 1
Calculate Initial PE
Total energy at top (KE = 0, all PE).
$E_{total} = mgh = m \times 9.8 \times 4.0 = 39.2m\text{ J}$
Step 2
Calculate Final KE
At the bottom, all energy is kinetic (plus losses).
$KE_f = \frac{1}{2}m(6)^2 = 18m\text{ J}$
Step 3
Find Percentage Lost
The difference is energy lost to friction. Note mass $m$ cancels.
$\text{Lost} = \frac{39.2m - 18m}{39.2m} \times 100\% = \frac{21.2}{39.2} \times 100\% \approx 54\%$
Problem Statement
A 70 kg man runs up stairs of vertical height 4.5 m in 4.0 seconds. What is his power output?
PowerGravitational PE
Strategic Toolkit
$$P = \frac{mgh}{t}$$
Step 1
Calculate Work Done
Work done against gravity equals the gain in gravitational PE.
$W = mgh = 70 \times 9.8 \times 4.5 = 3087\text{ J}$
Step 2
Compute Power
Divide work by time.
$P = \frac{3087}{4.0} = 771.75 \approx 770\text{ W}$
Problem Statement
Calculate the escape velocity from Earth's surface. ($R = 6.4 \times 10^6$ m, $g = 9.8$ m/s²)
Escape VelocityAbsolute PE
Strategic Toolkit
$$v_{esc} = \sqrt{2gR}$$
Step 1
Substitute Values
Use the simplified form with $g$ and $R$.
$v_{esc} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6}$
Step 2
Evaluate
Calculate the product under the square root.
$v_{esc} = \sqrt{1.2544 \times 10^8} = 1.12 \times 10^4\text{ m/s} \approx 11.2\text{ km/s}$
Memorize formulas
Work Done
Force times distance along the line of motion.
Formula
$W = Fd \cos \theta$
Work-Energy Theorem
Net work equals change in kinetic energy.
Formula
$W_{net} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$
Kinetic Energy
Energy of motion; depends on square of speed.
Formula
$KE = \frac{1}{2}mv^2$
Gravitational PE (near surface)
Energy from height in a uniform gravity field.
Formula
$U_g = mgh$
Elastic PE
Energy stored in a spring.
Formula
$U_s = \frac{1}{2}kx^2$
Absolute Gravitational PE
PE at distance r from Earth's center; zero at infinity.
Formula
$U = -\frac{GMm}{r}$
Escape Velocity
Minimum speed to leave a planet's gravitational field.
Formula
$v_{esc} = \sqrt{2gR} = \sqrt{\frac{2GM}{R}}$
Power
The rate of energy transfer.
Formula
$P = \frac{W}{t} = Fv$
Review
1
A force of 50 N is applied to a crate at an angle of 60° to the horizontal. If the crate moves 10 m horizontally, how much work is done by the force?
Ans
250 J
Review
Tap to reveal
Use the component of force along the displacement
Only the horizontal component of the applied force does work, so use the angle between the force and the displacement in the work formula.
$$W = Fd\cos\theta = 50 \times 10 \times \cos(60^\circ)$$
Evaluate the cosine term
Since $\cos 60^\circ = 0.5$, the work becomes half of $50 \times 10$.
$$W = 500 \times 0.5 = 250\ \text{J}$$
2
A car's speed triples from 10 m/s to 30 m/s. By what factor does its kinetic energy increase?
Ans
9
Review
Tap to reveal
Apply the speed-squared relationship
Kinetic energy is proportional to the square of speed, so any speed change must be squared before comparing energies.
$$\frac{KE_2}{KE_1} = \left(\frac{30}{10}\right)^2 = 3^2$$
Square the factor of three
Tripling the speed makes the kinetic energy nine times larger.
$$\frac{KE_2}{KE_1} = 9$$
3
A 500 kg cart starts from rest at the top of a 15 m high, 120 m long slope. If the average friction force along the slope is 300 N, what is its speed at the bottom?
Ans
12 m/s
Review
Tap to reveal
Find the available mechanical energy
The cart starts with gravitational potential energy, then loses some of it to friction along the full slope length.
$$PE = mgh = 500 \times 9.8 \times 15 = 73500\ \text{J}$$
$$W_f = fd = 300 \times 120 = 36000\ \text{J}$$
Convert the remaining energy into speed
Subtract the friction loss from the initial energy and set the remainder equal to kinetic energy at the bottom.
$$\frac{1}{2}mv^2 = 73500 - 36000 = 37500$$
$$v = \sqrt{\frac{2 \times 37500}{500}} \approx 12.2\ \text{m/s}$$
4
An engine performs 12,000 J of work in 2 minutes. What is the power output of the engine?
Ans
100 W
Review
Tap to reveal
Convert the time to seconds
Power in watts uses joules per second, so the 2-minute interval must be written in seconds first.
$$2\ \text{min} = 120\ \text{s}$$
Divide work by time
Use the definition of power to find the rate of doing work.
$$P = \frac{W}{t} = \frac{12000}{120} = 100\ \text{W}$$
5
A person holds a 20 kg box while walking 50 m on level ground. How much work does the holding force do on the box?
Ans
0 J
Review
Tap to reveal
Check the angle between force and motion
The holding force acts upward, but the box moves horizontally, so the force is perpendicular to the displacement.
$$\theta = 90^\circ$$
Use the zero-work case
A force perpendicular to displacement does no work because the cosine term becomes zero.
$$W = Fd\cos(90^\circ) = 0\ \text{J}$$
6
A spring with $k = 200$ N/m is compressed by 0.3 m. If compression is doubled to 0.6 m, the stored elastic PE becomes:
Ans
4 times the original
Review
Tap to reveal
Use the spring-energy formula
Elastic potential energy depends on the square of the compression distance, not just the distance itself.
$$U_s = \frac{1}{2}kx^2$$
Double the compression and square it
If the compression changes from $x$ to $2x$, the energy changes by the square of that factor.
$$\frac{U_2}{U_1} = \frac{(2x)^2}{x^2} = 4$$
7
The absolute gravitational potential energy of an object on Earth's surface is $-6.25 \times 10^7$ J/kg. The negative sign indicates that:
Ans
Energy must be supplied to free the object from Earth's gravitational field
Review
Tap to reveal
Recall the reference level for gravitational PE
If potential energy is defined to be zero at infinity, then objects inside Earth's gravitational field have negative potential energy.
$$U < 0\ \text{for a bound object}$$
Interpret the magnitude of the value
The size of the negative energy tells you how much energy must be added to move the object out of the gravitational well.
Energy must be supplied to free the object from Earth's gravitational field.
8
The escape velocity from Earth is about 11.2 km/s. If a planet has 3 times Earth's mass and the same radius, its escape velocity is closest to:
Ans
19.4 km/s
Review
Tap to reveal
Use the escape-velocity formula
Escape speed depends on the square root of the planet's mass when radius stays fixed.
$$v_{esc} = \sqrt{\frac{2GM}{R}}$$
Scale the mass and compare
Replacing Earth's mass by three times that value multiplies escape speed by $\sqrt{3}$.
$$v_{new} = \sqrt{3}\,(11.2) \approx 19.4\ \text{km/s}$$
9
A force of 400 N propels a car at a constant 80 km/h against road friction. The engine power must be:
Ans
8.9 kW
Review
Tap to reveal
Convert the speed to SI units
The power formula uses meters per second, so convert 80 km/h before substituting.
$$80\ \text{km/h} = 22.2\ \text{m/s}$$
Multiply force by speed
At constant speed, the engine power equals the friction force times the car's speed.
$$P = Fv = 400 \times 22.2 \approx 8889\ \text{W} \approx 8.9\ \text{kW}$$
10
A child starts from rest at the top of a 4.0 m slide. With friction, they reach the bottom at 6 m/s. The percentage of energy lost to friction is approximately:
Ans
54%
Review
Tap to reveal
Find the no-friction speed
Use the slide height to determine what speed the child would have if all potential energy became kinetic energy.
$$v_0 = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 4.0} \approx 8.85\ \text{m/s}$$
Compare actual and ideal kinetic energy
Kinetic energy is proportional to $v^2$, so the fraction retained is the ratio of squared speeds.
$$\frac{KE_{actual}}{KE_{ideal}} = \frac{6^2}{8.85^2} \approx 0.459$$
Convert the remainder to a loss percentage
Subtract the retained fraction from 1 to get the fraction lost to friction.
$$1 - 0.459 = 0.541 \approx 54\%$$
11
Two objects, one of mass $m$ and another of mass $2m$, are dropped from the same height on a frictionless track. At the bottom, their speeds are:
Ans
Equal — speed is independent of mass
Review
Tap to reveal
Set gravitational energy equal to kinetic energy
On a frictionless track, the lost gravitational potential energy becomes kinetic energy at the bottom.
$$mgh = \frac{1}{2}mv^2$$
Cancel the mass term
Because mass appears on both sides, it drops out and the speed depends only on height and gravity.
$$v = \sqrt{2gh}$$
Compare the two objects
Both objects fall through the same height, so both reach the same speed.
Equal speeds
12
A bullet of mass 10 g travelling at 400 m/s embeds in a 2 kg wooden block on a frictionless surface. Using conservation of momentum, the block's KE after collision is approximately:
Ans
4 J
Review
Tap to reveal
Conserve momentum for the inelastic collision
The bullet embeds in the block, so they move together afterward with a shared final speed.
$$0.01 \times 400 = 2.01\,v_f \Rightarrow v_f \approx 1.99\ \text{m/s}$$
Compute the post-collision kinetic energy
Use the combined mass and the shared speed to find the final kinetic energy of the system.
$$KE = \frac{1}{2}(2.01)(1.99)^2 \approx 4\ \text{J}$$
Question 1 of 12
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1
A force of 50 N is applied to a crate at an angle of 60° to the horizontal. If the crate moves 10 m horizontally, how much work is done by the force?