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

Grade 8IGCSE

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

🔑Concepts

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

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}}.

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.

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.

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

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).

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

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 0N0 \, 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=FforwardFfrictionF_{resultant} = F_{forward} - F_{friction}

💡Examples

Problem 1:

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

Solution:

v=st=450m30s=15m/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 25kg25 \, kg. Calculate its weight on Earth where g=10N/kgg = 10 \, N/kg.

Solution:

W=m×g=25kg×10N/kg=250NW = 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 50N50 \, N. The force of friction acting against the box is 15N15 \, N. Determine the resultant force.

Solution:

Fres=50N15N=35NF_{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.

Forces and Motion (Speed, Friction, and Gravity) Revision - Grade 8 Science IGCSE