derive P = Fv and use it to solve problems

Work, Energy & Power – Cambridge A‑Level Physics (9702)

1. Work

• Definition (vector form) \[ W = \int\limits_{C}\mathbf F\!\cdot\!d\mathbf s \] where the integral is taken along the path \(C\) of the displacement.
• Constant force \[ W = \mathbf F\!\cdot\!\mathbf s = Fs\cos\theta . \] \(\theta\) is the angle between \(\mathbf F\) and the displacement \(\mathbf s\).
• Units: joule (J) = newton·metre (N·m).
• Key points
  • \(\theta=0^{\circ}\) → \(W = Fs\) (maximum work).
  • \(\theta=90^{\circ}\) → \(W = 0\) (no work is done).

2. Work‑Energy Theorem

• The total work done by all forces acting on a particle equals the change in its kinetic energy: \[ \boxed{\Delta K = K_f-K_i = \sum W_{\text{all}} } . \]
• When the forces can be split into conservative (e.g. gravity, spring) and non‑conservative (e.g. friction, air drag) parts, \[ \sum W_{\text{all}} = \sum W_{\text{cons}} + \sum W_{\text{nc}} . \] The work of a conservative force can be expressed as the negative change in its associated potential energy, giving the familiar mechanical‑energy relation (see Section 4).

3. Kinetic and Gravitational Potential Energy

• Kinetic energy \[ K = \tfrac12 mv^{2}\qquad (m\; \text{mass},\; v\; \text{speed}) \]
• Gravitational potential energy (near Earth’s surface)

  Starting from the definition of the gravitational field, \( \mathbf g = \dfrac{\Delta\mathbf p}{\Delta h}\) (force per unit mass), the work done by gravity when an object of mass \(m\) moves through a vertical height \(\Delta h\) is

  \[ W_g = \int_{0}^{\Delta h} (-mg)\,dh = -mg\Delta h . \] Since \(W_g = -\Delta U_g\), the potential energy is \[ \boxed{U_g = mgh} \] with \(g = 9.81\;\text{m s}^{-2}\) and \(h\) measured from a chosen reference level.
• Both \(K\) and \(U_g\) are measured in joules (J).

4. Conservation of Mechanical Energy

For a system of particles the mechanical‑energy equation, derived from the work‑energy theorem, is

\[ K_i + U_i + W_{\text{nc}} = K_f + U_f . \]

• \(W_{\text{nc}}\) is the total work of all non‑conservative forces. Typical examples:
  • Friction on a horizontal surface: \(W_f = -f\,d\) (negative because the friction force opposes motion).
  • Air drag (quadratic form): \(W_{\text{drag}} = -\int F_{\text{drag}}\,ds\).
• If no non‑conservative forces act (\(W_{\text{nc}}=0\)), mechanical energy is conserved: \[ K_i + U_i = K_f + U_f . \]

5. Power

5.1 Instantaneous Power – Linear Motion

• Definition: \(P = \dfrac{dW}{dt}\).
• Using \(dW = \mathbf F\!\cdot\!d\mathbf s\) and \(d\mathbf s/dt = \mathbf v\): \[ \boxed{P = \mathbf F\!\cdot\!\mathbf v } . \]
• If \(\mathbf F\) is parallel to \(\mathbf v\) (the common exam case): \[ \boxed{P = Fv } . \]
• Units: watt (W) = joule s\(^{-1}\).

5.2 Average Power

\[ P_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t} \] Useful when the force or speed is constant over the interval \(\Delta t\).

5.3 Rotational Power

\[ \boxed{P = \tau \,\omega} \] where \(\tau\) is the torque (N·m) and \(\omega\) the angular velocity (rad s\(^{-1}\)).

5.4 Electrical Power

• Basic relation: \(P = VI\).
• Using Ohm’s law (\(V = IR\)): \[ P = I^{2}R = \frac{V^{2}}{R}. \]
• Internal resistance of a source (\(R_{\text{int}}\)): \[ P_{\text{loss}} = I^{2}R_{\text{int}} . \] The useful power delivered to the load is \(P_{\text{out}} = VI - I^{2}R_{\text{int}}\).
• AC circuits – when the supply is sinusoidal the real power is \[ P = VI\cos\phi , \] where \(\phi\) is the phase angle between voltage and current (power factor \(\cos\phi\)). This appears in Paper 3/5 questions on alternating currents.

5.5 Efficiency

\[ \boxed{\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times100\%} \] A high efficiency means most of the input power is converted into useful output.

6. Using \(P = Fv\) in Problem Solving

6.1 General Procedure

1. Read the question carefully – identify the system, the forces acting, and whether the motion is linear or rotational.
2. Determine if each force is constant or varies with time/position.
3. Write the instantaneous power for each force:
   • Linear: \(P = \mathbf F\!\cdot\!\mathbf v\) (or \(Fv\) when parallel).
   • Rotational: \(P = \tau\omega\).
   • Electrical: \(P = VI\) (include \(I^{2}R_{\text{int}}\) if internal resistance is mentioned).
4. If the problem asks for total work or energy over a time interval, integrate: \[ W = \int_{t_1}^{t_2} P\,dt . \]
5. Combine the power/energy results with the work‑energy theorem or the mechanical‑energy conservation equation (including any \(W_{\text{nc}}\) terms).
6. Solve algebraically for the required unknown (force, speed, power, distance, efficiency, etc.).

6.2 Worked Examples

Example 1 – Constant Horizontal Force (Linear)

Problem: A 2.0 kg block is pulled across a frictionless horizontal surface by a constant force \(F = 10\;\text{N}\). The block starts from rest. Find the instantaneous power when the block has moved 5 m.

Solution:

1. From the work‑energy theorem, \( \tfrac12 mv^{2}=Fd\). \[ v = \sqrt{\frac{2Fd}{m}} = \sqrt{\frac{2\times10\times5}{2}} = 7.07\;\text{m s}^{-1}. \]
2. Instantaneous power: \(P = Fv = 10\;\text{N}\times7.07\;\text{m s}^{-1}= 70.7\;\text{W}\) (≈ 71 W).

Example 2 – Variable Force (Linear)

Problem: A particle of mass \(m=0.5\;\text{kg}\) moves along the \(x\)-axis under a force \(F(x)=kx\) with \(k=4\;\text{N m}^{-1}\). It starts from rest at \(x=0\). Find (a) the power as a function of \(x\) and (b) the maximum power for \(0\le x\le2\;\text{m}\).

Solution:

1. Work from 0 to \(x\): \(W = \int_0^{x} kx'\,dx' = \tfrac12 kx^{2}\).
2. Set \(W = K\): \(\tfrac12 mv^{2}= \tfrac12 kx^{2}\) → \(v = x\sqrt{k/m}\).
3. Power: \(P(x)=F(x)v = (kx)\bigl[x\sqrt{k/m}\bigr]=kx^{2}\sqrt{k/m}\).
4. Insert numbers: \(P(x)=4x^{2}\sqrt{4/0.5}=8\sqrt{2}\,x^{2}\;\text{W}\).
5. Maximum at the upper limit \(x=2\;\text{m}\): \(P_{\max}=8\sqrt{2}\,(2)^{2}=32\sqrt{2}\approx45.3\;\text{W}\).

Example 3 – Real‑World: Motor Lifting a Load (Linear + Electrical)

Problem: A motor lifts a 20 kg load vertically at a constant speed of \(0.8\;\text{m s}^{-1}\). The motor is powered from a 240 V supply and has an efficiency of 85 %.

Required: (a) Mechanical power output, (b) Electrical power input, (c) The current drawn from the supply.

Solution:

1. Mechanical power (use \(P = Fv\) with \(F = mg\)): \[ P_{\text{out}} = mgv = (20\;\text{kg})(9.81\;\text{m s}^{-2})(0.8\;\text{m s}^{-1}) = 157\;\text{W}. \]
2. Electrical input from efficiency: \[ P_{\text{in}} = \frac{P_{\text{out}}}{\eta}= \frac{157}{0.85}=185\;\text{W}. \]
3. Current from \(P = VI\): \[ I = \frac{P_{\text{in}}}{V}= \frac{185\;\text{W}}{240\;\text{V}} = 0.77\;\text{A}. \]

Example 4 – Rotational Power

Problem: A uniform solid disc (mass \(M=2\;\text{kg}\), radius \(R=0.30\;\text{m}\)) is accelerated from rest by a constant torque \(\tau = 1.5\;\text{N m}\). Find the power delivered by the torque when the disc’s angular speed reaches \(\omega = 10\;\text{rad s}^{-1}\).

Solution:

1. Moment of inertia of a solid disc: \(I = \tfrac12 MR^{2}=0.09\;\text{kg m}^{2}\).
2. Instantaneous power: \(P = \tau\omega = 1.5\;\text{N m}\times10\;\text{rad s}^{-1}=15\;\text{W}\).

Example 5 – Electrical Power with Internal Resistance

Problem: A DC motor draws \(I = 8\;\text{A}\) from a 240 V supply. Its armature has an internal resistance of \(R_{\text{int}} = 0.5\;\Omega\). Determine the useful mechanical power if the motor’s efficiency (excluding internal losses) is 80 %.

Solution:

1. Electrical input: \(P_{\text{in}} = VI = 240\times8 = 1920\;\text{W}\).
2. Power lost as heat in the armature: \(P_{\text{loss}} = I^{2}R_{\text{int}} = 8^{2}\times0.5 = 32\;\text{W}\).
3. Power available to the motor’s output shaft: \(P_{\text{avail}} = P_{\text{in}}-P_{\text{loss}} = 1888\;\text{W}\).
4. Mechanical output (including the given efficiency): \[ P_{\text{out}} = \eta P_{\text{avail}} = 0.80\times1888 = 1510\;\text{W}. \]

7. Summary Table of Key Relations

Quantity	Symbol	Expression	Units
Work	\(W\)	\(W = \displaystyle\int\mathbf F\!\cdot\!d\mathbf s = \mathbf F\!\cdot\!\mathbf s\) (constant \(F\))	J (N m)
Work‑energy theorem		\(\Delta K = \sum W_{\text{all}}\)	J
Kinetic energy	\(K\)	\(K = \tfrac12 mv^{2}\)	J
Gravitational potential energy	\(U_g\)	\(U_g = mgh\)	J
Mechanical‑energy conservation		\(K_i+U_i+W_{\text{nc}} = K_f+U_f\)	J
Linear instantaneous power	\(P\)	\(P = \mathbf F\!\cdot\!\mathbf v\) (scalar: \(P = Fv\) if parallel)	W
Average power	\(P_{\text{avg}}\)	\(P_{\text{avg}} = \dfrac{W}{\Delta t}\)	W
Rotational power	\(P\)	\(P = \tau\omega\)	W
Electrical power (ideal)	\(P\)	\(P = VI = I^{2}R = V^{2}/R\)	W
Electrical power (with internal resistance)		\(P_{\text{loss}} = I^{2}R_{\text{int}}\), \(P_{\text{out}} = VI - I^{2}R_{\text{int}}\)	W
Electrical power (AC, real)		\(P = VI\cos\phi\) (power factor \(\cos\phi\))	W
Efficiency	\(\eta\)	\(\displaystyle\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\times100\%\)	%

8. Common Pitfalls

• Average vs. instantaneous power – Use \(P_{\text{avg}} = W/t\) only when force and speed are constant. Otherwise apply \(P = \mathbf F\!\cdot\!\mathbf v\) at each instant.
• Neglecting the dot product – Only the component of the force parallel to the velocity contributes to power.
• Omitting non‑conservative work – Friction, air drag, and any applied force that does not store energy must appear as \(W_{\text{nc}}\) (or as a negative term like \(-fd\)).
• Mixing linear and rotational formulas – Remember \(P = Fv\) for translation, \(P = \tau\omega\) for rotation, and \(P = VI\) for electricity.
• Sign conventions – Positive work adds energy to the system; negative work removes energy (e.g., friction).
• Electrical power in real circuits – Always check whether internal resistance or power factor is mentioned; otherwise the simple \(VI\) formula will give an over‑estimate.

9. Practice Questions

1. A 1.5 kg cart is accelerated from rest by a constant horizontal force of 12 N over a distance of 3 m. Find the instantaneous power supplied by the force when the cart has moved 2 m.
2. A cyclist exerts a constant pedal force of 150 N on the crank, which moves with a linear speed of 0.5 m s\(^{-1}\). What is the instantaneous power output? (Assume the force is tangent to the circular motion of the crank.)
3. A 0.8 kg particle slides down a frictionless incline of height 4 m. Determine the power due to gravity when the particle’s speed is 6 m s\(^{-1}\).
4. A motor lifts a 20 kg load vertically at a constant speed of 0.8 m s\(^{-1}\). The motor’s efficiency is 85 %. Calculate the electrical power drawn from a 240 V supply.
5. A solid cylinder (mass \(M=5\) kg, radius \(R=0.10\) m) is spun up from rest by a constant torque of 0.8 N m. Find the power delivered by the torque when the angular speed reaches 30 rad s\(^{-1}\).

10. Suggested Diagram