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Physics - Forces and Motion

Grade 7Cambridge (IGCSE)

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

🔑Concepts

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Distance is a scalar quantity representing the total path length, while displacement is a vector quantity representing the change in position in a specific direction.

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Speed is defined as the distance traveled per unit of time (v=dtv = \frac{d}{t}) and is measured in m/sm/s.

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Velocity is speed in a specified direction. It changes if either the speed or the direction of motion changes.

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Acceleration is the rate of change of velocity over time, measured in m/s2m/s^2. It is calculated using a=v−uta = \frac{v - u}{t}.

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A force is a push or a pull acting upon an object resulting from its interaction with another object. Forces are measured in Newtons (NN).

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Newton's Second Law states that the acceleration of an object is directly proportional to the resultant force and inversely proportional to its mass: F=m×aF = m \times a.

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Mass is the amount of matter in an object (measured in kgkg), whereas weight is the force of gravity acting on that mass (W=m×gW = m \times g).

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Friction is a force that opposes the motion of an object. It always acts in the opposite direction to the direction of motion.

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Balanced forces result in a zero resultant force, meaning an object at rest stays at rest, and an object in motion continues at a constant velocity.

📐Formulae

v=dtv = \frac{d}{t}

a=v−uta = \frac{v - u}{t}

F=m×aF = m \times a

W=m×gW = m \times g

Resultant Force=∑F\text{Resultant Force} = \sum F

💡Examples

Problem 1:

A car travels a distance of 240 meters240\text{ meters} in a time of 12 seconds12\text{ seconds}. Calculate its average speed.

Solution:

v=240 m12 s=20 m/sv = \frac{240\text{ m}}{12\text{ s}} = 20\text{ m/s}

Explanation:

To find the speed, we use the formula v=dtv = \frac{d}{t}, substituting the given distance and time.

Problem 2:

An object with a mass of 5 kg5\text{ kg} is pushed with a resultant force of 15 N15\text{ N}. What is the acceleration of the object?

Solution:

a=Fm=15 N5 kg=3 m/s2a = \frac{F}{m} = \frac{15\text{ N}}{5\text{ kg}} = 3\text{ m/s}^2

Explanation:

Using Newton's Second Law F=m×aF = m \times a, we rearrange the formula to solve for acceleration: a=Fma = \frac{F}{m}.

Problem 3:

On Earth, where the gravitational field strength gg is approximately 9.8 N/kg9.8\text{ N/kg}, calculate the weight of a 10 kg10\text{ kg} suitcase.

Solution:

W=m×g=10 kg×9.8 N/kg=98 NW = m \times g = 10\text{ kg} \times 9.8\text{ N/kg} = 98\text{ N}

Explanation:

Weight is the product of mass and the gravitational acceleration constant (gg).