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Space, Time and Motion - Work, Energy and Power

Grade 12IBPhysics

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

🔑Concepts

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Work done WW is defined as the product of the magnitude of the displacement ss and the component of the force FF in the direction of the displacement: W=Fscos⁡θW = Fs \cos \theta. It is a scalar quantity measured in Joules (JJ).

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

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Gravitational Potential Energy (ΔEp\Delta E_p) is the energy stored in an object due to its position in a gravitational field: ΔEp=mgΔh\Delta E_p = mg\Delta h.

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Elastic Potential Energy (EelE_{el}) is the energy stored as a result of deforming an elastic object: Eel=12kΔx2E_{el} = \frac{1}{2}k\Delta x^2.

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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, the total energy remains constant.

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Work-Energy Theorem: The net work done on an object is equal to its change in kinetic energy: Wnet=ΔEkW_{net} = \Delta E_k.

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Power (PP) is the rate at which work is done or energy is transferred: P=ΔWΔtP = \frac{\Delta W}{\Delta t}. For an object moving at constant velocity vv under a force FF, P=FvP = Fv.

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

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The area under a Force-Displacement graph represents the work done.

📐Formulae

W=Fscos⁡θW = Fs \cos \theta

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

ΔEp=mgΔh\Delta E_p = mg\Delta h

Eel=12kΔx2E_{el} = \frac{1}{2}k\Delta x^2

P=ΔWΔtP = \frac{\Delta W}{\Delta t}

P=FvP = Fv

η=PoutPin\eta = \frac{P_{out}}{P_{in}}

💡Examples

Problem 1:

A block of mass 2.0 kg2.0 \text{ kg} is pulled 5.0 m5.0 \text{ m} along a horizontal surface by a force of 10 N10 \text{ N} acting at an angle of 30∘30^\circ to the horizontal. If the coefficient of dynamic friction is 0.150.15, calculate the net work done on the block.

Solution:

  1. Horizontal component of applied force: Fx=10cos⁡30∘≈8.66 NF_x = 10 \cos 30^\circ \approx 8.66 \text{ N}.
  2. Normal force: R=mg−Fsin⁡30∘=(2.0×9.81)−(10×0.5)=19.62−5.0=14.62 NR = mg - F \sin 30^\circ = (2.0 \times 9.81) - (10 \times 0.5) = 19.62 - 5.0 = 14.62 \text{ N}.
  3. Frictional force: f=μR=0.15×14.62≈2.19 Nf = \mu R = 0.15 \times 14.62 \approx 2.19 \text{ N}.
  4. Net force: Fnet=8.66−2.19=6.47 NF_{net} = 8.66 - 2.19 = 6.47 \text{ N}.
  5. Net Work: W=Fnet×s=6.47×5.0=32.35 JW = F_{net} \times s = 6.47 \times 5.0 = 32.35 \text{ J}.

Explanation:

We resolve the applied force into components, calculate the normal reaction force to find friction, determine the resultant force in the direction of motion, and multiply by displacement.

Problem 2:

An electric motor with an efficiency of 75%75\% is used to lift a 50 kg50 \text{ kg} mass vertically at a constant speed of 2.0 m s−12.0 \text{ m s}^{-1}. Calculate the electrical power input to the motor.

Solution:

  1. Useful power output needed to lift the mass: Pout=Fv=mgvP_{out} = Fv = mgv.
  2. Pout=50×9.81×2.0=981 WP_{out} = 50 \times 9.81 \times 2.0 = 981 \text{ W}.
  3. Using efficiency formula: η=PoutPin⇒0.75=981Pin\eta = \frac{P_{out}}{P_{in}} \Rightarrow 0.75 = \frac{981}{P_{in}}.
  4. Pin=9810.75=1308 WP_{in} = \frac{981}{0.75} = 1308 \text{ W} or 1.31 kW1.31 \text{ kW}.

Explanation:

First, calculate the mechanical power required to overcome gravity at constant speed, then use the efficiency factor to find the higher electrical power required as input.

Work, Energy and Power - Revision Notes & Key Formulas | IB Grade 12 Physics