Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.

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Q. Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.

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Concept understanding — Power Against Gravity

Power Against Gravity: The Intuition

Imagine you're lifting a bucket of water from a well. The bucket is heavy — gravity is pulling it down with a force equal to its weight. To lift it, you must apply an upward force that exactly cancels gravity. If you lift it slowly, you feel the strain for a long time. If you yank it up quickly, you feel a burst of effort, but it's over fast.

That "burst of effort per unit time" is what we call power. When you're working against a constant force like gravity, power tells you how fast you're doing that work.

Note

Power is not the force itself, nor the work alone — it's the rate at which work is done. Lifting the same bucket to the same height requires the same total work, regardless of speed. But doing it faster requires more power.

The Precise Statement

When an object moves vertically against gravity (upward), the force you must supply is at least equal to the object's weight:

F = mg

where $m$ is mass and $g$ is acceleration due to gravity ( $9.8\ \text{m/s}^2$ on Earth).

The work done to lift it through a height $h$ is:

W = F \cdot h = mgh

Now, power $P$ is work per unit time. If you lift it in time $t$ :

P = \frac{W}{t} = \frac{mgh}{t}

But $\frac{h}{t}$ is just the upward speed $v$ (assuming constant speed). So we get the compact form:

P = mg \cdot v

This is power against gravity — the rate at which you must supply energy to lift a mass $m$ at constant speed $v$ against the pull of gravity.

What This Really Means

It's a minimum. If you accelerate the object upward, you need even more force (Newton's second law), and hence more power. The formula $P = mgv$ assumes you're lifting at steady speed — no acceleration.
Direction matters. If the object moves downward at constant speed, gravity does the work, and you (or a brake) must absorb power. The formula still gives the magnitude, but the sign flips.
It's independent of path. Only the vertical speed matters. Whether you lift straight up or along a ramp, the power against gravity depends only on the vertical component of velocity.

Tip

For a quick calculation: lifting a 10 kg mass at 0.5 m/s requires $P = 10 \times 9.8 \times 0.5 \approx 49$ watts. That's about the power of a dim incandescent bulb — and you'd feel it after a minute.

Common Mistake to Avoid …

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Power is the rate of doing work. Lifting $m=100\,\text{kg}$ through $h=10\,\text{m}$ needs work $W=mgh$ , done in $t=20\,\text{s}$ , so $P=\dfrac{mgh}{t}=490\,\text{W}$ .

Concept: power as work per unit time

The crane raises the load against gravity, increasing its gravitational potential energy by $mgh$ . Power is the rate at which this work is done.

P=\frac{W}{t}=\frac{mgh}{t}

Step-by-step solution

1. Work done against gravity.

W=mgh=100\times 9.8\times 10=9800\,\text{J}

2. Power.

$P=\frac{W}{t}=\frac{9800}{20}=490\,\text{W}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2024Set D-21 markMCQ
$t = 5~\mathrm{s}$
›Reveal solutionSolution

Power is the rate of doing work, so the area under the P–t graph is the kinetic energy gained; convert that energy to momentum with $p=\sqrt{2mK}$ .

1. Link power to energy

By definition,
$P = \frac{dW}{dt} \quad\Longrightarrow\quad W = \int_0^{t} P\,dt = \text{area under the } P\text{–}t \text{ graph}$
2. Work–energy theorem

The particle starts from rest ( $K_i = 0$ ) and moves along a straight line, so all the work delivered goes into kinetic energy:
$W = \Delta K = K_f - 0 = K_f$
3. Read the graph

The graph is a straight line through the origin, i.e. $P = kt$ , rising to $P = 10\ \mathrm{W}$ at $t = 5\ \mathrm{s}$ (so $k = 2\ \mathrm{W\,s^{-1}}$ ). The area under it up to $t = 5\ \mathrm{s}$ is a triangle:
$W = \frac{1}{2}\times \text{base} \times \text{height} = \frac{1}{2}\times 5\ \mathrm{s} \times 10\ \mathrm{W} = 25\ \mathrm{J}$
(Equivalently $\int_0^5 2t\,dt = t^2\big|_0^5 = 25\ \mathrm{J}$ .)

4. Kinetic energy → momentum

For a particle of mass $m$ ,

$K = \frac{p^2}{2m} \quad\Longrightarrow\quad p = \sqrt{2mK}$ …
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COMEDK 2024Set 2024-E1 markMCQ
$80 \mathrm{~m}$
›Reveal solutionSolution
The turbine’s output power is the gravitational potential energy per second of the falling water, reduced by 20% frictional losses. The result is 12.8 kW, which corresponds to option (A).

The core idea here is simple: water falling from a height loses gravitational potential energy, which is converted into kinetic energy and then into mechanical work by the turbine. But not all of that energy is usable — friction in the water flow and the turbine mechanism wastes a fixed percentage. So we calculate the input power (the rate at which potential energy is delivered) and then subtract the losses.

Why this approach works:

Power is energy per second. The water’s mass flow rate (kg/s) and the height (m) give us the rate of potential energy change: $P_{\text{input}} = \dot{m} g h$ . Then we apply the efficiency (or loss factor) to find the actual electrical/mechanical power output.

Identify the given data

Height: $h = 80 \, \text{m}$

Mass flow rate: $\dot{m} = 20 \, \text{kg/s}$

Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$ (standard value)

Frictional loss: 20% of input energy → usable energy = 80% of input.

Calculate the input power (rate of potential energy)

The potential energy of a mass $m$ at height $h$ is $mgh$ . Per second, this becomes:

$P_{\text{input}} = \dot{m} g h = 20 \times 9.8 \times 80$
First, $20 \times 9.8 = 196$ . Then $196 \times 80 = 15680 \, \text{W}$ .

So $P_{\text{input}} = 15680 \, \text{W} = 15.68 \, \text{kW}$ .

Account for the frictional losses Losses are 20%, so the turbine receives only 80% of the input power: $P_{\text{output}} = 0.80 \times 15.68 \, \text{kW} = 12.544 \, \text{kW}$ …
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Q.Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.

Concept understanding — Power Against Gravity

Power Against Gravity: The Intuition

The Precise Statement

What This Really Means

Common Mistake to Avoid …

Concept: power as work per unit time

Step-by-step solution

Concept: Power as the Rate of Doing Work

1. Link power to energy

2. Work–energy theorem

3. Read the graph

4. Kinetic energy → momentum

Q.Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.

Concept: power as work per unit time

Step-by-step solution

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