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Physics - Forces and Motion (Speed, Friction, and Gravity)

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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Speed is a scalar quantity defined as the distance traveled per unit of time, measured in meters per second (m/sm/s).

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Average speed is calculated by taking the total distance traveled and dividing it by the total time taken: vavg=dtotalttotalv_{avg} = \frac{d_{total}}{t_{total}}.

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Friction is a contact force that opposes the motion of an object. It acts in the opposite direction to the direction of movement or intended movement.

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Air resistance (drag) is a type of friction that occurs when an object moves through the air. It increases as the speed of the object increases.

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Mass (mm) is the amount of matter in an object and is measured in kilograms (kgkg). It remains constant regardless of location.

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Weight (WW) is the force of gravity acting on an object's mass, measured in Newtons (NN). Weight changes depending on the gravitational field strength (gg).

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On Earth, the gravitational field strength (gg) is approximately 9.8 N/kg9.8 \, N/kg, though often rounded to 10 N/kg10 \, N/kg for IGCSE Grade 8 calculations.

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A resultant force is the single force that has the same effect as all the forces acting on an object. If forces are balanced, the resultant force is 0 N0 \, N.

📐Formulae

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

s=v×ts = v \times t

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

W=m×gW = m \times g

Fresultant=Fforward−FfrictionF_{resultant} = F_{forward} - F_{friction}

💡Examples

Problem 1:

A cyclist travels a distance of 450 m450 \, m in 30 s30 \, s. Calculate the average speed of the cyclist.

Solution:

v=st=450 m30 s=15 m/sv = \frac{s}{t} = \frac{450 \, m}{30 \, s} = 15 \, m/s

Explanation:

To find the speed, divide the total distance by the time taken. The unit is m/sm/s.

Problem 2:

An object has a mass of 25 kg25 \, kg. Calculate its weight on Earth where g=10 N/kgg = 10 \, N/kg.

Solution:

W=m×g=25 kg×10 N/kg=250 NW = m \times g = 25 \, kg \times 10 \, N/kg = 250 \, N

Explanation:

Weight is the product of mass and gravitational field strength. The resulting unit is Newtons (NN).

Problem 3:

A box is pushed across a floor with a forward force of 50 N50 \, N. The force of friction acting against the box is 15 N15 \, N. Determine the resultant force.

Solution:

Fres=50 N−15 N=35 NF_{res} = 50 \, N - 15 \, N = 35 \, N

Explanation:

Since friction opposes motion, we subtract the frictional force from the applied forward force to find the net (resultant) force acting on the box.