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Physics - Motion, Forces and Energy

Grade 10IGCSE

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

🔑Concepts

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Speed is defined as the distance traveled per unit time (v=dtv = \frac{d}{t}), whereas velocity is a vector quantity representing speed in a given direction.

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Acceleration aa is the rate of change of velocity. On a velocity-time graph, the gradient represents acceleration and the area under the graph represents the distance traveled.

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Newton's Second Law states that the resultant force FF acting on an object is equal to the product of its mass mm and acceleration aa (F=maF = ma).

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Mass is the amount of matter in an object and remains constant, while Weight WW is the force of gravity acting on that mass (W=mgW = mg).

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Hooke's Law states that the extension xx of a spring is directly proportional to the force FF applied, provided the limit of proportionality is not exceeded (F=kxF = kx).

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The Moment of a force is its turning effect about a pivot, calculated as Force ×\times perpendicular distance from the pivot (M=F×dM = F \times d).

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Work is done when a force moves an object through a distance (W=FdW = Fd). Power PP is the rate at which work is done or energy is transferred (P=WtP = \frac{W}{t}).

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The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred from one form to another. Efficiency is the ratio of useful energy output to total energy input.

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Pressure PP is the force exerted per unit area (P=FAP = \frac{F}{A}). In liquids, pressure increases with depth hh and density ρ\rho (P=ρghP = \rho gh).

📐Formulae

v=stv = \frac{s}{t}

a=v−uta = \frac{v - u}{t}

F=maF = ma

W=mgW = mg

F=kxF = kx

Moment=F×d\text{Moment} = F \times d

P=FAP = \frac{F}{A}

p=ρghp = \rho gh

W=FdW = Fd

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

ΔEp=mgh\Delta E_p = mgh

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

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

💡Examples

Problem 1:

A car of mass 1500 kg1500\text{ kg} accelerates uniformly from rest to a velocity of 20 m/s20\text{ m/s} in 5 s5\text{ s}. Calculate the resultant force acting on the car.

Solution:

First, find the acceleration: a=v−ut=20−05=4 m/s2a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4\text{ m/s}^2. Then, use Newton's Second Law: F=ma=1500×4=6000 NF = ma = 1500 \times 4 = 6000\text{ N}.

Explanation:

We first determine the rate of change of velocity (acceleration) and then apply the force formula to find the required resultant force.

Problem 2:

A ball of mass 0.5 kg0.5\text{ kg} is held at a height of 10 m10\text{ m} above the ground. Calculate its Gravitational Potential Energy (EpE_p) and its velocity just before hitting the ground (assume g=9.8 m/s2g = 9.8\text{ m/s}^2 and no air resistance).

Solution:

Ep=mgh=0.5×9.8×10=49 JE_p = mgh = 0.5 \times 9.8 \times 10 = 49\text{ J}. By conservation of energy, Ek=EpE_k = E_p, so 12mv2=49\frac{1}{2}mv^2 = 49. Solving for vv: v=2×490.5=196=14 m/sv = \sqrt{\frac{2 \times 49}{0.5}} = \sqrt{196} = 14\text{ m/s}.

Explanation:

The potential energy at the top is completely converted into kinetic energy at the bottom, allowing us to calculate the final velocity.

Problem 3:

Calculate the pressure exerted by a block of weight 200 N200\text{ N} resting on a surface with a base area of 0.5 m20.5\text{ m}^2.

Solution:

P=FA=2000.5=400 PaP = \frac{F}{A} = \frac{200}{0.5} = 400\text{ Pa}.

Explanation:

Pressure is the distribution of force over a specific area. Dividing the weight (force) by the contact area gives the pressure in Pascals (1 Pa=1 N/m21\text{ Pa} = 1\text{ N/m}^2).

Motion, Forces and Energy - Revision Notes & Key Formulas | IGCSE Grade 10 Science