Work, Energy, and Power

Work (CED Unit 4)

Work is the energy transferred to or from an object by a force acting through a displacement.

Definition

For a constant force $\vec{F}$acting through displacement $\vec{d}$:

$$W = \vec\{F\} \cdot \vec\{d\} = Fd\cos\theta$$

Where $\theta$ is the angle between the force and the displacement.

• $\theta = 0^\circ$: $W = Fd$ (force in direction of motion)
• $\theta = 90^\circ$: $W = 0$ (force perpendicular to motion)
• $\theta = 180^\circ$: $W = -Fd$ (force opposes motion)

Why the Dot Product

Work is a scalar, not a vector. It is the component of the force along the direction of motion That does work. The perpendicular component changes the direction of motion but does not transfer Energy. This is why the normal force does no work on an object sliding along a surface, and why the Centripetal force does no work on an object in uniform circular motion.

Work by a Variable Force (AP Physics C)

$$W = \int_a^b F(x)\, dx$$

The work done by a variable force equals the area under the force-vs-displacement curve.

Units

1 Joule (J) = $1 \mathrm{ N \cdot \mathrm{m}} = 1 \mathrm{ kg \cdot \mathrm{m^2/\mathrm{s^2}}}$

Dimensional Analysis of the Joule

From $W = Fd$: $[W] = \mathrm\{N \cdot \mathrm\{m = (\mathrm\{kg \cdot \mathrm\{m/s^2) \cdot \mathrm\{m = \mathrm\{kg \cdot \mathrm\{m^2/\mathrm\{s^2$. This is the same as the dimensions of kinetic energy $\frac{1}{2}mv^2$ Confirming that work and Energy are the same physical quantity.

Power at Constant Speed Up an Incline

At constant speed, the driving force equals the component of weight down the slope plus friction:

$$P = F_\{\mathrm\{drive\} v = (mg\sin\theta + f)v$$

Potential Energy Diagrams and Stability

A potential energy diagram plots $U(x)$ vs $x$. From this, we can determine:

• Force: $\displaystyle F(x) = -\frac{dU}{dx}$
• Equilibrium points: Where $F(x) = 0$ I.e., $\frac{dU}{dx} = 0$
• Stable equilibrium: Local minimum of $U$ ($\frac{d^2U}{dx^2} \gt 0$)
• Unstable equilibrium: Local maximum of $U$ ($\frac{d^2U}{dx^2} \lt 0$)
• Neutral equilibrium: Flat region of $U$

Why $F = -dU/dx$

Consider a one-dimensional conservative force $F(x)$. The work done by this force from $x_1$ to $x_2$ is:

$$W = \int_\{x_1\}^\{x_2\} F(x)\, dx$$

By definition, the potential energy change is the negative of this work:

$$\Delta U = U(x_2) - U(x_1) = -W = -\int_\{x_1\}^\{x_2\} F(x)\, dx$$

Differentiating with respect to $x_2$:

$$\frac\{dU\}\{dx\} = -F(x)$$

Total Energy on a Potential Energy Diagram

On a potential energy diagram, the total energy $E = K + U$ is a horizontal line. The kinetic energy at any point is $K = E - U$. Motion is only possible where $K \ge 0$ I.e., $E \ge U$. The object Oscillates between the “turning points” where $E = U$ (and $K = 0$).

Escape Velocity

The minimum speed needed for an object to escape a gravitational field (reach infinity with zero Speed):

$$\frac\{1\}\{2\}mv_e^2 - \frac\{GMm\}\{R\} = 0 \implies v_e = \sqrt\{\frac\{2GM\}\{R\}\}$$

For Earth: $v_e \approx 11.2 \mathrm{ km/s}$.

Why Escape Velocity Is Independent of Mass

Both the kinetic energy $\frac{1}{2}mv_e^2$ and the gravitational potential energy $GMm/R$ are Proportional to $m$ So the mass cancels. A ping-pong ball and a spacecraft need the same escape Velocity from the same point. This is a consequence of the equivalence of gravitational and inertial Mass.

Intuition

Work, energy, and power provide a scalar alternative to vector-based force analysis. Instead of tracking forces and accelerations, you track energy, a single number that captures the state of the system. The work-energy theorem is the bridge: net work done on an object equals its change in kinetic energy.

Conservation of energy intuition: Energy cannot be created or destroyed, only transformed. In an isolated system, the total energy is constant. This is more powerful than Newton’s second law whenever forces change direction or magnitude during the motion: energy is a scalar (no components to track), and the work-energy theorem integrates the varying force for you - no component-wise integration required.ble or unknown.

Power intuition: Power is the rate of energy transfer, how fast you do work. Two cars can deliver the same work (same KE change), but the sports car does it in less time (higher power). On the AP exam, problems often ask “how long does it take?” which is a power question, not an energy question.

Non-conservative forces: Friction and air resistance dissipate energy as heat. When non-conservative forces act, mechanical energy is not conserved, you must account for the energy lost to heat. The work-energy theorem still applies, but you include the work done by non-conservative forces.

Common Pitfalls

1. Forgetting that work is a scalar. It can be positive, negative, or zero depending on the angle between force and displacement.
2. Using the wrong height for gravitational PE. $h$ is measured from the chosen reference level, which can be set anywhere convenient.
3. Confusing force and power. Power is force times velocity, not just force.
4. Applying conservation of mechanical energy when non-conservative forces are present. Use $W_{\mathrm{nc}} = \Delta E$ instead.
5. Forgetting the sign convention. Work done by gravity is $W_g = -mg\Delta h$ (positive when the object moves down).
6. Using $U = mgh$ far from Earth’s surface. Use $U = -\frac{GMm}{r}$ instead.
7. Incorrectly computing power as force divided by time. Power is work divided by time, or force times velocity.
8. Confusing the total energy with the mechanical energy. Total energy includes thermal energy, sound energy, and all other forms. Mechanical energy ($K + U$) is conserved only when non-conservative forces do no work.

Practice Questions

1. A $5.0 \mathrm{ kg}$ block slides $3.0 \mathrm{ m}$ down a $25^\circ$ frictionless incline. Find the work done by gravity and the final speed.

2. A spring with $k = 500 \mathrm{ N/m}$ is compressed $0.10 \mathrm{ m}$ and launches a $0.50 \mathrm{ kg}$ block on a frictionless surface. Find the maximum speed of the block and the maximum height it reaches on a frictionless incline.

3. A $1000 \mathrm{ kg}$ elevator starts from rest and accelerates upward at $2.0 \mathrm{ m/s^2}$ for $3.0 \mathrm{ s}$. Find the power output of the motor at $t = 3.0 \mathrm{ s}$.

4. A pendulum of length $1.5 \mathrm{ m}$ is released from horizontal. Find its speed at the lowest point (neglect air resistance).

5. A $0.50 \mathrm{ kg}$ ball is thrown straight up with speed $15 \mathrm{ m/s}$. If air resistance does $-5.0 \mathrm{ J}$ of work, find the maximum height.

6. A car of mass $1500 \mathrm{ kg}$ travels at constant speed $20 \mathrm{ m/s}$ up a $10^\circ$ incline. If the engine produces $50 \mathrm{ kW}$ Find the frictional force.

7. A satellite in circular orbit at altitude $500 \mathrm{ km}$ needs to escape Earth. What additional speed must it acquire?

8. Derive the relationship $F = -\frac{dU}{dx}$ from the work-energy theorem for one-dimensional motion.

9. A force $F(x) = 3x^2 - 2x$ (in N, with $x$ in m) acts on a $2.0 \mathrm{ kg}$ object. Find the work done as the object moves from $x = 0$ to $x = 3.0 \mathrm{ m}$ And the speed at $x = 3.0$ m if the object started from rest.

10. A potential energy function is given by $U(x) = x^4 - 2x^2$ (in J, with $x$ in m). Find all equilibrium positions and classify each as stable, unstable, or neutral.

11. Work by a Variable Force: Detailed Examples (AP Physics C)

Worked Example: Spring Force

A spring obeys Hooke’s law with $k = 200 \mathrm{ N/m}$. Find the work done compressing it from $x = 0$ to $x = 0.15 \mathrm{ m}$.

$$W = \int_0^{0.15} (-kx)\, dx = -\left[\frac{1}{2}kx^2\right]_0^{0.15} = -\frac{1}{2}(200)(0.0225) = -2.25 \mathrm{ J}$$

The negative sign indicates the spring force opposes the displacement. The magnitude of work done by the spring is $2.25 \mathrm{ J}$ And the work done on the spring (by the external agent) is $+2.25 \mathrm{ J}$ Which is stored as elastic potential energy.

Worked Example: Non-Linear Force

A force $F(x) = 3x^2 + 2x$ (in N, with $x$ in m) acts on a $2 \mathrm{ kg}$ object. Find the work Done from $x = 0$ to $x = 3 \mathrm{ m}$ and the speed at $x = 3 \mathrm{ m}$ if the object starts from rest.

$$W = \int_0^3 (3x^2 + 2x)\, dx = \left[x^3 + x^2\right]_0^3 = 27 + 9 = 36 \mathrm{ J}$$

By the work-energy theorem: $W = \frac{1}{2}mv_f^2 - 0$

$$v_f = \sqrt{\frac{2W}{m}} = \sqrt{\frac{72}{2}} = \sqrt{36} = 6 \mathrm{ m/s}$$

12. Conservation of Energy: Extended Examples

Worked Example: Block-Spring System on a Frictionless Surface

A $3 \mathrm{ kg}$ block slides on a frictionless surface at $4 \mathrm{ m/s}$ and strikes a spring with $k = 500 \mathrm{ N/m}$. Find the maximum compression of the spring.

At maximum compression, all kinetic energy has been converted to elastic potential energy:

$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$ $$x = v\sqrt{\frac{m}{k}} = 4\sqrt{\frac{3}{500}} = 4 \times 0.0775 = 0.310 \mathrm{ m}$$

Worked Example: Pendulum with Energy

A pendulum of length $L = 1.5 \mathrm{ m}$ and mass $m = 0.5 \mathrm{ kg}$ is released from Horizontal ($\theta = 90^{\circ}$). Find the speed at the lowest point.

The pendulum bob falls a height $h = L = 1.5 \mathrm{ m}$.

$$mgh = \frac{1}{2}mv^2$$ $$v = \sqrt{2gL} = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{29.4} = 5.42 \mathrm{ m/s}$$

Worked Example: Loop-the-Loop

A block of mass $m$ slides from rest down a frictionless ramp of height $h$ and enters a vertical Circular loop of radius $R$. What is the minimum $h$ for the block to complete the loop?

At the top of the loop, the minimum speed to maintain contact is $v = \sqrt{gR}$ (when the normal Force is zero).

Using conservation of energy from the top of the ramp to the top of the loop (height $= 2R$):

$$mgh = \frac{1}{2}mv^2 + mg(2R) = \frac{1}{2}m(gR) + 2mgR = \frac{5}{2}mgR$$ $$h = \frac{5}{2}R = 2.5R$$

The ramp must be at least 2.5 times the radius of the loop.

13. Potential Energy Diagrams: Extended Analysis

Worked Example: Interpreting $U(x) = x^4 - 2x^2$

$$F(x) = -\frac{dU}{dx} = -(4x^3 - 4x) = -4x(x^2 - 1) = -4x(x-1)(x+1)$$

Equilibrium points: $F(x) = 0$ at $x = 0, \pm 1$.

Classification:

$$\frac{d^2U}{dx^2} = 12x^2 - 4$$

• At $x = 0$: $\frac{d^2U}{dx^2} = -4 \lt 0$ — unstable equilibrium (local maximum)
• At $x = \pm 1$: $\frac{d^2U}{dx^2} = 8 \gt 0$ — stable equilibrium (local minima)

$U(0) = 0$ (unstable), $U(\pm 1) = 1 - 2 = -1$ (stable, energy $= -1$ J).

An object with total energy $E = -0.5$ J oscillates between the two turning points in one of the Potential wells. It cannot escape because $E \lt U(0) = 0$.

14. Power: Extended Analysis

Instantaneous Power for Variable Force

When force and velocity are not constant:

$$P(t) = \vec{F}(t) \cdot \vec{v}(t)$$

Worked Example: Power as a Function of Time

A force $F(t) = 6t$ N acts on a $2 \mathrm{ kg}$ object initially at rest. Find the power at $t = 3$ S.

$$a = \frac{F}{m} = 3t$$ $$v(t) = \int_0^t 3t'\, dt' = \frac{3t^2}{2}$$ $$P(3) = F(3) \cdot v(3) = 18 \times \frac{27}{2} = 18 \times 13.5 = 243 \mathrm{ W}$$

Worked Example: Power Needed to Climb at Constant Speed

A $70 \mathrm{ kg}$ cyclist climbs a $6^{\circ}$ hill at $5 \mathrm{ m/s}$. The total resistive Force (friction plus air resistance) is $20 \mathrm{ N}$. Find the power output.

$$P = Fv = (mg\sin\theta + F_{\mathrm{resist})v} = (70 \times 9.8 \times \sin 6^{\circ} + 20) \times 5$$ $$= (70 \times 9.8 \times 0.1045 + 20) \times 5 = (71.7 + 20) \times 5 = 458.5 \mathrm{ W}$$

This is close to the maximum sustained power output of a trained cyclist, illustrating the Difficulty of climbing even a modest gradient.

15. Summary Table: Energy Forms and Equations

16. Common Pitfalls: Extended

1. Using the wrong sign for work done by gravity. When an object moves down, gravity does positive work ($W_g = +mg\Delta h$ Where $\Delta h$ is negative if up is positive). When an object moves up, gravity does negative work. Be careful with sign conventions.

2. Confusing the reference level for gravitational PE. The choice of $h = 0$ is arbitrary. What matters is the change in height, $\Delta h$ Not the absolute height.

3. Applying $U = mgh$ when the height change is a significant fraction of the distance from Earth’s centre. Use $U = -GMm/r$ for large altitude changes.

4. Forgetting that power is the time derivative of work, not force. $P = dW/dt = Fv$. Power is not $F/t$ or $F/t^2$.

5. Assuming the force is constant when using $W = Fd\cos\theta$. If the force varies with position, you must integrate: $W = \int F(x)\, dx$.

Practice Questions (Additional)

1. A $1.5 \mathrm{ kg}$ object slides from rest down a curved frictionless ramp of height $3 \mathrm{ m}$ and enters a rough horizontal section with $\mu_k = 0.4$. How far does it slide on the rough section before stopping?

2. A spring with $k = 800 \mathrm{ N/m}$ is compressed $0.10 \mathrm{ m}$ and launches a $0.2 \mathrm{ kg}$ ball vertically. Find the maximum height reached (a) on Earth and (b) on the Moon ($g = 1.62 \mathrm{ m/s^2}$).

3. A force $F(x) = -kx + bx^3$ acts on a particle. Find the potential energy function $U(x)$ and identify the equilibrium positions.

4. A $50 \mathrm{ kg}$ child on a swing is pushed to a height $0.8 \mathrm{ m}$ above the lowest point. Find the speed at the lowest point and the speed when the swing is $0.4 \mathrm{ m}$ above the lowest point.

5. The potential energy of a particle is $U(r) = \frac{A}{r} - \frac{B}{r^2}$ where $A$ and $B$ are positive constants. Find (a) the force, (b) the equilibrium position, and (c) whether the equilibrium is stable or unstable.

Extended Worked Examples

Example 16: Power Dissipated by Air Resistance

A car of mass $1500 \mathrm{ kg}$ travels at a constant $25 \mathrm{ m/s}$ on a level road. The Engine produces a driving force of $1200 \mathrm{ N}$. Calculate the power output of the engine and The power dissipated by air resistance and rolling friction.

Step 1: At constant speed, net force is zero

$$F_{\mathrm{drive}} = F_{\mathrm{resist}} = 1200 \mathrm{ N}$$

Step 2: Power output

$$P = Fv = 1200 \times 25 = 30000 \mathrm{ W} = 30 \mathrm{ kW}$$

Step 3: All power goes to overcoming resistance

$$P_{\mathrm{dissipated}} = 30 \mathrm{ kW}$$

If the car then accelerates to $30 \mathrm{ m/s}$ with the same driving force:

Step 4: New resistance at $30 \mathrm{ m/s}$ (assuming air resistance scales as $v^2$)

$$F_{\mathrm{air} \propto v^2}$$

Let $F_{\mathrm{air,25}} = F_{\mathrm{air}(25)}$ and $F_{\mathrm{roll}}$ be the constant rolling Friction.

At $25 \mathrm{ m/s}$: $F_{\mathrm{air}(25)} + F_{\mathrm{roll}} = 1200 \mathrm{ N}$

At $30 \mathrm{ m/s}$: $F_{\mathrm{air}(30)} = F_{\mathrm{air}(25) \times (30/25)^2} = 1.44 \times F_{\mathrm{air}(25)}$

Assuming $F_{\mathrm{air}(25)} = 1000 \mathrm{ N}$ and $F_{\mathrm{roll}} = 200 \mathrm{ N}$:

$$F_{\mathrm{air}(30)} = 1440 \mathrm{ N}$$ $$F_{\mathrm{total resist}(30)} = 1440 + 200 = 1640 \mathrm{ N}$$

Net force: $F_{\mathrm{net}} = 1200 - 1640 = -440 \mathrm{ N}$

The car cannot maintain $30 \mathrm{ m/s}$ with the same driving force — air resistance increases Too rapidly.

Example 17: Energy in a Mass-Spring System with Gravity

A $2 \mathrm{ kg}$ block hangs from a vertical spring with $k = 200 \mathrm{ N/m}$. The block is Pulled down $0.15 \mathrm{ m}$ from equilibrium and released. Find the maximum speed and the maximum Height above the release point.

Step 1: Find the equilibrium extension

At equilibrium, spring force balances weight:

$$kx_0 = mg \implies x_0 = \frac{mg}{k} = \frac{2 \times 9.8}{200} = 0.098 \mathrm{ m}$$

Step 2: Define the zero of potential energy

The simplest approach: measure displacements from equilibrium. Gravity shifts the equilibrium Position but does not affect the oscillation frequency or amplitude.

Step 3: Maximum speed (at equilibrium position)

Using energy conservation relative to equilibrium:

$$\frac{1}{2}kA^2 = \frac{1}{2}mv_{\max}^2$$

Where $A = 0.15 \mathrm{ m}$ is the amplitude.

$$v_{\max} = A\sqrt{\frac{k}{m}} = 0.15 \times \sqrt{\frac{200}{2}} = 0.15 \times 10 = 1.5 \mathrm{ m/s}$$

Step 4: Maximum height above release point

The block oscillates symmetrically about equilibrium, so it rises $0.15 \mathrm{ m}$ above Equilibrium to $0.15 \mathrm{ m}$ above equilibrium, which is $0.15 + 0.15 = 0.30 \mathrm{ m}$ above the lowest point (the release point).

Example 18: Work Done Against a Non-Conservative Force

A $5 \mathrm{ kg}$ block starts from rest at the top of a rough curved ramp of height $3 \mathrm{ m}$. At the bottom, its speed is $6 \mathrm{ m/s}$. Calculate the work done by friction.

Step 1: Energy at the top

$$E_{\mathrm{top}} = mgh = 5 \times 9.8 \times 3 = 147 \mathrm{ J}$$

Step 2: Energy at the bottom

$$E_{\mathrm{bottom}} = \frac{1}{2}mv^2 = \frac{1}{2} \times 5 \times 36 = 90 \mathrm{ J}$$

Step 3: Work done by friction

$$W_f = E_{\mathrm{bottom}} - E_{\mathrm{top}} = 90 - 147 = -57 \mathrm{ J}$$

The friction does $-57 \mathrm{ J}$ of work (dissipates $57 \mathrm{ J}$ of mechanical energy as Heat).

Step 4: Effective coefficient of friction (if the ramp were straight)

If the ramp length is $L = 6 \mathrm{ m}$:

$$W_f = -fL = -\mu mg\cos\theta \times L$$ $$57 = \mu \times 5 \times 9.8 \times \cos\theta \times 6$$

With $\cos\theta = 3/6 = 0.5$ (from $h = 3 \mathrm{ m}$, $L = 6 \mathrm{ m}$):

$$57 = \mu \times 5 \times 9.8 \times 0.5 \times 6 = 147\mu$$ $$\mu = \frac{57}{147} = 0.388$$

Common Pitfalls Extended

Pitfall 6: Confusing Power and Energy

Power is the rate of energy transfer ($P = dW/dt$), measured in watts (W). Energy is the total Transferred, measured in joules (J). A device with high power can deliver the same energy as a Low-power device in less time. Always check units: if a question asks for energy, the answer must be in joules; if it asks for power, the answer must be in watts.

Pitfall 7: Forgetting That Work Is a Scalar

Work is $W = \vec{F} \cdot \vec{d} = Fd\cos\theta$. It can be positive, negative, or zero, but it has no direction. A common error is to assign a direction to work or to add work components Vectorially instead of algebraically.

Pitfall 8: Incorrect Reference Points for Gravitational PE

Gravitational PE depends on the choice of zero point. Only changes in PE are physically Meaningful: $\Delta PE = mg\Delta h$. Always be explicit about where you are measuring $h$ from, and Ensure you are consistent throughout the problem.

Additional Practice Problems

1. A $1200 \mathrm{ kg}$ car accelerates from $0$ to $30 \mathrm{ m/s}$ in $8 \mathrm{ s}$. Calculate (a) the average power developed, (b) the instantaneous power at $t = 4 \mathrm{ s}$ assuming constant acceleration, and (c) the total distance covered.

2. A $0.5 \mathrm{ kg}$ ball is dropped from a height of $20 \mathrm{ m}$. On each bounce, it loses $20\%$ of its kinetic energy. Calculate the height after the first bounce, the velocity just before the second bounce, and the total vertical distance travelled before the ball comes to rest.

3. A force $F = 10 + 3x$ (N) acts on a $4 \mathrm{ kg}$ object from $x = 0$ to $x = 5 \mathrm{ m}$. Calculate the work done, the final speed if starting from rest, and the power delivered at $x = 3 \mathrm{ m}$ if the object is moving at $4 \mathrm{ m/s}$ at that point.

4. A roller coaster car of mass $500 \mathrm{ kg}$ starts from rest at point A, height $30 \mathrm{ m}$. It goes through a loop of radius $10 \mathrm{ m}$ and then up a ramp to point B. If $80\%$ of the initial PE is converted to KE at the top of the loop, calculate the normal force at the top of the loop and the maximum possible height of point B.

5. Two springs with $k_1 = 200 \mathrm{ N/m}$ and $k_2 = 300 \mathrm{ N/m}$ are connected (a) in series and (b) in parallel. For each arrangement, find the effective spring constant, the period of oscillation for a $2 \mathrm{ kg}$ mass, and the energy stored when compressed $0.1 \mathrm{ m}$.

Practice Problems

Question 1: Work-energy theorem with variable force

A $3 \mathrm{ kg}$ object moves along the x-axis under the influence of a force $F(x) = 4x^2 - 2x \mathrm{ N}$ (where $x$ is in metres). Calculate the work done by this force as The object moves from $x = 0$ to $x = 3 \mathrm{ m}$ and the final speed if it started from rest.

Answer

$W = \int_0^3 F(x) \, dx = \int_0^3 (4x^2 - 2x) \, dx = \left[\frac{4x^3}{3} - x^2\right]_0^3 = \frac{4(27)}{3} - 9 = 36 - 9 = 27 \mathrm{ J}$.

By the work-energy theorem: $W = \Delta KE = \frac{1}{2}mv^2 - 0$.

$v = \sqrt{2W/m} = \sqrt{2 \times 27/3} = \sqrt{18} = 4.24 \mathrm{ m/s}$.

Question 2: Conservation of energy with spring

A $0.5 \mathrm{ kg}$ block slides on a frictionless surface and collides with a horizontal spring ($k = 200 \mathrm{ N/m}$). The block compresses the spring by $0.15 \mathrm{ m}$ before momentarily Stopping. What was the speed of the block just before it hit the spring?

Answer

All kinetic energy converts to elastic potential energy: $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$.

$v = x\sqrt{k/m} = 0.15\sqrt{200/0.5} = 0.15\sqrt{400} = 0.15 \times 20 = 3.0 \mathrm{ m/s}$.

Question 3: Power and inclined plane

A $60 \mathrm{ kg}$ person runs up a flight of stairs that is $10 \mathrm{ m}$ high in $8 \mathrm{ s}$. Calculate the average power output in watts and horsepower. Take $g = 9.8 \mathrm{ m/s^2}$ and $1 \mathrm{ hp} = 746 \mathrm{ W}$.

Answer

Work done against gravity: $W = mgh = 60 \times 9.8 \times 10 = 5880 \mathrm{ J}$.

Average power: $P = W/t = 5880/8 = 735 \mathrm{ W}$.

In horsepower: $735/746 = 0.985 \mathrm{ hp}$.

Question 4: Loop-the-loop energy analysis

A block of mass $m$ starts from rest at height $h$ on a frictionless track and enters a circular Loop of radius $R$. What is the minimum value of $h$ for the block to complete the loop? Express your answer in terms of $R$.

Answer

At the top of the loop, the minimum condition is that the normal force is zero, so centripetal Acceleration is provided entirely by gravity: $mg = mv_{\mathrm{top}^2/R}$ Giving $v_{\mathrm{top}^2} = gR$.

Energy conservation from start to top of loop (height $2R$):

$mgh = mg(2R) + \frac{1}{2}mv_{\mathrm{top}^2} = 2mgR + \frac{1}{2}m(gR) = \frac{5}{2}mgR$.

$h = \frac{5}{2}R = 2.5R$.

Question 5: Work done by non-constant force on a curve

A $2 \mathrm{ kg}$ object is moved from the origin to the point $(4, 3)$ by a force $\vec{F} = (3x\hat{i} + 2y\hat{j}) \mathrm{ N}$. Calculate the work done by this force.

Answer

$W = \int \vec{F} \cdot d\vec{r} = \int_0^4 3x \, dx + \int_0^3 2y \, dy$.

$= \left[\frac{3x^2}{2}\right]_0^4 + \left[y^2\right]_0^3 = 24 + 9 = 33 \mathrm{ J}$.

The force is conservative (it can be written as the negative gradient of a potential $U = -\frac{3}{2}x^2 - y^2$), so the work depends only on the endpoints, not the path taken.

Cross-References

• Kinematics: Work-energy theorem connects force and displacement to changes in speed without needing time.
• Momentum and Impulse: Work-energy and impulse-momentum are complementary approaches, energy for scalar problems, momentum for vector problems.
• Newton’s Laws: Newton’s second law is the foundation; work-energy is its integral form.
• AP Calculus, Integrals: Work is defined as the integral of force over displacement, making integration the mathematical foundation of energy analysis.