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Power Is F · v: Use It for Instantaneous Values and the Constant-Power Pattern

Power is the rate of doing work. Questions use the average value W/t, the instantaneous value F · v, or the constant-power motion in which speed and acceleration follow a fixed pattern in time.

NSEJS · Work Energy and Power · 14 linked questions · 100% coverage

Know this first

Power and Rate of Doing Work is the central idea in this question pattern. Identify the condition in the stem before applying the method below.

Study deeper: NCERT Science · Work Energy and Power

Average and instantaneous power

P_avg = W/t and P = F · v

For a 1 kg body under

F = 2t i + 3t² j, the velocity is v = t² i + t³ j, so P = 2t³ + 3t⁵

The tension in a rope lifting a 10 kg block at g/2 is 150 N, and over 4 s the block rises 40 m, so the average power of the tension is 150 x 40 / 4 = 1500 W.

For a continuous source of water falling a height h,

P = (mass per second) g h: 100 kg/s from 100 m gives 100 kW

For a hose, the kinetic energy imparted per second is (1/2) (mass per unit length) v³, and a heart that pumps 10 litres a minute at a pressure of 75 mm of mercury (10 200 Pa) gives P = p x flow rate = 10 200 x 1.67 x 10⁻⁴ = 1.7 W.

Constant-power motion

With P constant,

P = F v = m v (dv/dt) gives v² = 2 P t / m, so v is proportional to t^(1/2), a is proportional to t^(-1/2) and F = sqrt(P m / (2t))

a = v dv/dx gives m v² dv = P dx, so m v³/3 = P x and x is proportional to v³

These patterns identify a constant-power machine in a graph or an equation.

Efficiency, pumps and units

Required input power is useful power divided by efficiency: a 4 kg body moving up an incline of 1 in 40 at 40 m/s needs F v = 4 x 9.8 x (1/40) x 40 = 39.2 W of useful power, so at 50% efficiency 78.4 W. To get n times the water through the same pipe in the same time, the speed must rise n times; since force equals mass flow rate times speed (ρ A v x v), the force rises n² times. One horse power is about 746 W: a 1 hp motor lifting 2238 kg through 30 m (20 m of well plus 10 m to the tank) needs 2238 x 10 x 30 / 746, close to 900 s, or 15 minutes.

OlympIQ exam lens

✓P = W/t on average; P = F · v at an instant
✓Constant power: v is proportional to t^(1/2), a to t^(-1/2), x to v³
✓Water falling: P = (mass per second) x g x h; hose: (1/2) x (mass per length) x v³
✓Input power = useful power / efficiency
✓1 hp is about 746 W; include the full lift height from the source

OlympIQ trap

Do not mix average and instantaneous power. A machine delivering constant power has a changing force, and the force-times-velocity product at one instant is not the work divided by a long interval unless the power is constant.

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