P = dW/dt = F·v. Car engine power problem.

Power = Force × Velocity — Derivation and Application

Question

A car of mass 1500 kg moves on a straight road at a constant speed of 20 m/s. The frictional force acting on the car is 3000 N. Find the power delivered by the car’s engine.

(CBSE 2024 Board Exam)

Solution — Step by Step

At constant speed, acceleration = 0. By Newton’s first law, net force = 0. So the engine thrust must exactly equal the friction force — the engine provides 3000 N of force to keep moving.

Power is rate of doing work:

P = \frac{dW}{dt}

Since work done by a constant force is $W = F \cdot s$ , differentiating with respect to time:

P = F \cdot \frac{ds}{dt} = F \cdot v

This is the key formula: P = Fv.

P = F \times v = 3000 \times 20 = 60{,}000 \text{ W}

P = 60 kW

Why This Works

The formula $P = Fv$ comes directly from the definition of power. Work is force times displacement, and power is how fast that work happens. When you divide $F \cdot \Delta s$ by $\Delta t$ , the $\Delta s / \Delta t$ naturally becomes velocity.

This is why a car struggles more to maintain the same speed uphill — it needs to supply more force (against gravity + friction), which means higher power at the same velocity. The engine’s rated power is its ceiling; once you hit it, speed can no longer increase.

In JEE and NEET numericals, “constant velocity” is a signal to equate driving force with resistive force. The problem is telling you the engine force without saying so directly.

Alternative Method — Using Energy

Instead of $P = Fv$ , we can think in terms of energy dissipation.

Since the car moves at constant speed, all engine work goes into overcoming friction (converted to heat). In one second, the car travels 20 m. Work done against friction in that time:

W = F \times d = 3000 \times 20 = 60{,}000 \text{ J}

Since this happens in 1 second:

P = \frac{W}{t} = \frac{60{,}000}{1} = 60{,}000 \text{ W} = 60 \text{ kW}

Same answer, different path. The first method is faster for exams; this one builds better physical intuition.

Common Mistake

Students often try to use $P = \frac{1}{2}mv^2 / t$ — mixing up kinetic energy with work done by the engine. Kinetic energy is constant here (constant speed), so $\Delta KE = 0$ . The engine’s work goes entirely to fighting friction, not building up KE. Using KE in this formula gives a wrong and dimensionally inconsistent answer.

The second trap: forgetting to check whether “engine force = friction force.” If the question said the car was accelerating, you’d need to find the net force first, then add friction back to get engine force. Always sketch the force diagram before writing $P = Fv$ .

P = \frac{W}{t} \quad \text{(average power)}

P = Fv \quad \text{(instantaneous power, F parallel to v)}

P = Fv\cos\theta \quad \text{(F at angle } \theta \text{ to velocity)}

1 kW = 1000 W, 1 HP ≈ 746 W

This question is a 2-mark scoring topic in CBSE boards and appears as a single-step calculation in JEE Main. Once you see “constant speed + friction given,” the answer is one line: $P = Fv$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Energy

Common Mistake

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