Home / iGCSE Physics (0625) 1.5.1 Effects of force Paper 3 -Exam Style Questions- New Syllabus

iGCSE Physics (0625) 1.5.1 Effects of force Paper 3 -Exam Style Questions- New Syllabus

Question

A girl is cycling along a straight horizontal road.
Fig. 1.1 shows the directions of the forces acting on the cyclist as she cycles in the direction of
force C.

(a) State which force shows the direction of:
    (i) the force due to gravity
    (ii) the force due to air resistance.
    (iii) Force A changes and becomes larger than force C. State any effect this change has on the motion of the cyclist.

(b) Another cyclist travels a distance of 250 m in a time of 21 s.
    (i) Calculate the average speed of the cyclist.
    (ii) The cyclist exerts a force of 36N to move the cycle forwards. Calculate the work done by this force when the cyclist travels 250 m. Include the unit.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):

Topic 1.5.1 — Effects of forces (Parts (a)(i), (a)(ii), (a)(iii))
Topic 1.2 — Motion (Part (b)(i))
Topic 1.7.2 — Work (Part (b)(ii))

▶️ Answer/Explanation

(a)(i) D
The force due to gravity (weight) always acts vertically downwards towards the centre of the Earth. In the diagram, force D is the only arrow pointing straight down.

(a)(ii) A
Air resistance (drag) is a frictional force that opposes the motion of an object moving through the air. Since the cyclist moves in the direction of force C, force A points backwards and represents the air resistance.

(a)(iii) decelerating / slowing down / less speed
Force C is the forward driving force, and force A is the backward resistive force. When force A becomes larger than force C, the resultant force acts opposite to the direction of motion, causing the cyclist to decelerate.

(b)(i) 12 (m/s)
Average speed is calculated by dividing the total distance travelled by the total time taken: $v = \frac{s}{t} = \frac{250\text{ m}}{21\text{ s}} = 11.9\text{ m/s}$, which rounds to $12\text{ m/s}$.

(b)(ii) 9000 J
Work done is the product of force and the distance moved in the direction of the force: $W = F \times d = 36\text{ N} \times 250\text{ m} = 9000\text{ J}$. The unit of work is the joule (J).

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