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Space, Time and Motion - Work, Energy and Power

Grade 11IBPhysics

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

🔑Concepts

Work done WW is defined as the product of the force FF and the displacement ss in the direction of the force: W=FscosθW = Fs \cos \theta, where θ\theta is the angle between the force and the displacement vector.

Work is a scalar quantity measured in Joules (JJ). On a force-displacement graph, the work done is represented by the area under the curve.

Kinetic Energy EkE_k is the energy an object possesses due to its motion, given by Ek=12mv2E_k = \frac{1}{2}mv^2.

Gravitational Potential Energy ΔEp\Delta E_p is the energy stored in an object due to its vertical position in a gravitational field, calculated as ΔEp=mgΔh\Delta E_p = mg\Delta h.

Elastic Potential Energy EhE_h is stored in a deformed elastic object (like a spring) and is given by Eel=12kx2E_{el} = \frac{1}{2}kx^2 provided the limit of proportionality is not exceeded.

The Principle of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. In an isolated system, Etotal=Ek+Ep+Q=constantE_{total} = E_k + E_p + Q = \text{constant}, where QQ represents internal energy (heat).

Power PP is the rate at which work is done or the rate at which energy is transferred, measured in Watts (WW). It can also be expressed as the product of force and velocity: P=FvP = Fv.

Efficiency η\eta is the ratio of useful energy (or power) output to the total energy (or power) input: η=useful outputtotal input\eta = \frac{\text{useful output}}{\text{total input}}.

📐Formulae

W=FscosθW = Fs \cos \theta

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

ΔEp=mgΔh\Delta E_p = mg\Delta h

Eel=12kx2E_{el} = \frac{1}{2}kx^2

P=ΔWΔtP = \frac{\Delta W}{\Delta t}

P=FvP = Fv

Efficiency(η)=WoutWin=PoutPin\text{Efficiency} (\eta) = \frac{W_{out}}{W_{in}} = \frac{P_{out}}{P_{in}}

💡Examples

Problem 1:

A block of mass 2.0 kg2.0\text{ kg} is pulled along a horizontal frictionless surface by a force of 15 N15\text{ N} acting at an angle of 3030^\circ to the horizontal. Calculate the work done by the force in moving the block a distance of 5.0 m5.0\text{ m}.

Solution:

W=FscosθW = Fs \cos \theta W=15×5.0×cos(30)W = 15 \times 5.0 \times \cos(30^\circ) W=75×0.86665 JW = 75 \times 0.866 \approx 65\text{ J}

Explanation:

Work is only done by the component of the force that acts in the direction of the displacement (Fcos30F \cos 30^\circ). The vertical component does no work as there is no vertical displacement.

Problem 2:

An electric motor with an efficiency of 75%75\% is used to lift a 50 kg50\text{ kg} crate vertically at a constant speed of 2.0 m s12.0\text{ m s}^{-1}. Calculate the electrical power input required for the motor.

Solution:

First, find the useful power output: Pout=Fv=mgvP_{out} = Fv = mgv Pout=50×9.81×2.0=981 WP_{out} = 50 \times 9.81 \times 2.0 = 981\text{ W} Next, use the efficiency formula: η=PoutPin    0.75=981Pin\eta = \frac{P_{out}}{P_{in}} \implies 0.75 = \frac{981}{P_{in}} Pin=9810.75=1308 W1.3 kWP_{in} = \frac{981}{0.75} = 1308\text{ W} \approx 1.3\text{ kW}

Explanation:

The motor must provide enough power to overcome gravity at the given velocity. Since the motor is not perfectly efficient, the input electrical power must be greater than the mechanical power output.

Work, Energy and Power - Revision Notes & Key Formulas | IB Grade 11 Physics