derive P = Fv and use it to solve problems

Energy Conservation

In a closed system the total mechanical energy (kinetic + potential) remains constant unless acted upon by external non‑conservative forces. This principle allows us to relate forces, motions and the rate at which energy is transferred – the concept of power.

Derivation of the Power Formula $P = Fv$

Consider a particle of mass $m$ moving in a straight line under a constant force $F$ in the direction of motion.

1. Work done by the force over a small displacement $dx$ is $dW = F\,dx$.
2. By definition, power is the rate at which work is done: $$P = \frac{dW}{dt}$$
3. Substituting $dW = F\,dx$ gives $$P = \frac{F\,dx}{dt}$$
4. Since $dx/dt$ is the instantaneous velocity $v$, we obtain $$\boxed{P = Fv}$$

Key points:

• If the force is not parallel to the velocity, only the component of the force in the direction of motion contributes: $P = \mathbf{F}\cdot\mathbf{v}$.
• The formula is valid for any instantaneous values of $F$ and $v$, even when they vary with time.

Using $P = Fv$ in Problem Solving

When a problem provides two of the three quantities (force, velocity, power), the third can be found directly. Often the challenge is to combine this relation with energy‑conservation ideas.

General Procedure

1. Identify the system and list all forces acting on it.
2. Determine whether the forces are constant or varying. If varying, express them as functions of time or displacement.
3. Write the power expression $P = \mathbf{F}\cdot\mathbf{v}$ for each relevant force.
4. If the problem involves a time interval, integrate power to obtain work or energy: $$W = \int_{t_1}^{t_2} P\,dt$$
5. Apply the conservation of mechanical energy: $$K_i + U_i + W_{\text{nc}} = K_f + U_f$$ where $W_{\text{nc}}$ is the work done by non‑conservative forces (often expressed via $P$).
6. Solve for the unknown quantity.

Example Problem

Problem: A 2.0 kg block is pulled across a horizontal frictionless surface by a constant horizontal force of 10 N. The block starts from rest. Determine the power delivered by the force when the block has moved 5 m.

Solution:

1. Since the surface is frictionless, the only horizontal force is $F = 10\,$N.
2. Use energy conservation to find the speed after moving 5 m: $$\frac{1}{2}mv^2 = F\,d \quad\Rightarrow\quad v = \sqrt{\frac{2Fd}{m}} = \sqrt{\frac{2\times10\times5}{2}} = \sqrt{50}=7.07\ \text{m s}^{-1}$$
3. Apply $P = Fv$: $$P = (10\ \text{N})(7.07\ \text{m s}^{-1}) = 70.7\ \text{W}$$

Thus the power supplied by the pulling force at the instant the block has travelled 5 m is approximately 71 W.

Another Example – Variable Force

Problem: A particle of mass $m = 0.5\,$kg moves along the $x$‑axis under a force $F(x)=kx$ where $k = 4\,$N m$^{-1}$. The particle starts from rest at $x=0$. Find the power as a function of $x$ and the maximum power attained before the particle reaches $x=2\,$m.

Solution:

1. Force as a function of position: $F(x)=kx$.
2. Work done from 0 to $x$: $$W = \int_0^x kx'\,dx' = \frac{1}{2}kx^2$$
3. Kinetic energy equals work (no other forces): $$\frac{1}{2}mv^2 = \frac{1}{2}kx^2 \;\Rightarrow\; v = x\sqrt{\frac{k}{m}}$$
4. Power: $$P(x)=F(x)v = (kx)\left[x\sqrt{\frac{k}{m}}\right] = kx^2\sqrt{\frac{k}{m}}$$
5. Insert numbers: $k=4\,$N m$^{-1}$, $m=0.5\,$kg, $$P(x)=4x^2\sqrt{\frac{4}{0.5}} = 4x^2\sqrt{8}=4x^2\,(2\sqrt{2}) = 8\sqrt{2}\,x^2\ \text{W}$$
6. At $x=2\,$m, $$P_{\max}=8\sqrt{2}\,(2)^2 = 8\sqrt{2}\times4 = 32\sqrt{2}\approx45.3\ \text{W}$$

Summary Table of Key Relations

Quantity	Symbol	Expression	Units
Force	$F$	Newton's second law $F = ma$ (or given)	newton (N)
Velocity	$v$	Derivative of displacement, $v = \frac{dx}{dt}$	metre per second (m s$^{-1}$)
Power	$P$	$P = \mathbf{F}\cdot\mathbf{v}$ (scalar $P = Fv$ when parallel)	watt (W)
Kinetic Energy	$K$	$K = \frac{1}{2}mv^2$	joule (J)
Potential Energy (gravitational)	$U_g$	$U_g = mgh$	joule (J)
Work	$W$	$W = \int \mathbf{F}\cdot d\mathbf{s}$	joule (J)

Common Pitfalls

• Confusing average power ($W/t$) with instantaneous power ($P = Fv$). Use the latter when the force or velocity changes during the interval.
• Neglecting the vector nature of the dot product. If the force is not parallel to the motion, only the component along the velocity contributes.
• For rotating systems, replace $Fv$ with torque $\tau$ times angular velocity $\omega$: $P = \tau\omega$.

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. Calculate the 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 crank’s circular motion.)
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$^{-1}$.

These notes provide the theoretical foundation and a systematic method for applying the power formula $P = Fv$ in a variety of A‑Level physics contexts.