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Physics - Energy, Work, and Power

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Work is done when a force FF acts on an object and causes a displacement ss. It is a scalar quantity measured in Joules (JJ).

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Energy is the capacity to do work. Like work, it is measured in Joules (JJ).

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Kinetic Energy (EkE_k) is the energy of an object due to its motion. It depends on the mass mm and the square of the velocity vv.

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Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its vertical position or height hh in a gravitational field gg.

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The Law of Conservation of Energy states that in an isolated system, the total energy remains constant. Energy can be transformed from one form to another but cannot be created or destroyed: Einitial=EfinalE_{initial} = E_{final}.

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Power (PP) is the rate at which work is done or the rate at which energy is transferred. It is measured in Watts (WW), where 1W=1Js−11 W = 1 J s^{-1}.

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Efficiency is the ratio of useful energy output to the total energy input, often expressed as a percentage.

📐Formulae

W=Fscos⁡θW = Fs \cos \theta

Ek=12mv2E_k = \frac{1}{2}mv^2

Ep=mghE_p = mgh

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

P=FvP = Fv

Efficiency=Useful Work OutputTotal Energy Input×100%\text{Efficiency} = \frac{\text{Useful Work Output}}{\text{Total Energy Input}} \times 100\%

💡Examples

Problem 1:

A 0.5kg0.5 kg ball is thrown vertically upwards with an initial velocity of 15ms−115 m s^{-1}. Calculate the maximum height reached by the ball, ignoring air resistance (use g=9.8ms−2g = 9.8 m s^{-2}).

Solution:

Using conservation of energy: Ek(bottom)=Ep(top)E_k(\text{bottom}) = E_p(\text{top}). Thus, 12mv2=mgh\frac{1}{2}mv^2 = mgh. Solving for hh: h=v22g=1522×9.8=22519.6≈11.48mh = \frac{v^2}{2g} = \frac{15^2}{2 \times 9.8} = \frac{225}{19.6} \approx 11.48 m.

Explanation:

At the maximum height, all initial kinetic energy is converted into gravitational potential energy. The mass mm cancels out from both sides of the equation.

Problem 2:

An electric crane lifts a load of 2000kg2000 kg to a height of 20m20 m in 40s40 s. Calculate the power developed by the crane.

Solution:

W=mgh=2000×9.8×20=392,000JW = mgh = 2000 \times 9.8 \times 20 = 392,000 J. P=Wt=392,00040=9,800W=9.8kWP = \frac{W}{t} = \frac{392,000}{40} = 9,800 W = 9.8 kW.

Explanation:

First, calculate the work done against gravity (mghmgh), then divide by the time taken to find the rate of work (power).

Problem 3:

A box is pushed 5m5 m across a floor by a horizontal force of 20N20 N. If the work done is 100J100 J, but the friction force is 5N5 N, calculate the efficiency of the process in moving the box.

Solution:

Total work input Win=20N×5m=100JW_{in} = 20 N \times 5 m = 100 J. Net force Fnet=20N−5N=15NF_{net} = 20 N - 5 N = 15 N. Useful work Wout=15N×5m=75JW_{out} = 15 N \times 5 m = 75 J. Efficiency=75100×100%=75%\text{Efficiency} = \frac{75}{100} \times 100\% = 75\%.

Explanation:

Efficiency is calculated by comparing the useful work (the work done by the net force) to the total work input provided by the applied force.

Energy, Work, and Power - Revision Notes & Key Formulas | IB Grade 10 Science