iGCSE Physics (0625) 1.7.2 Work Paper 3 -Exam Style Questions- New Syllabus
Questions
Fig. 4.1 shows a student using a battery-powered device.

(a) State the energy store in the battery.
(b) The student pushes the device along the floor at a constant speed with a horizontal force of 14N. The student pushes the device for a distance of 4.5m. Calculate the mechanical work done by the force pushing the device.
(c) The student uses the device for a time of 30s. The energy input to the device is 5400J. Calculate the power input to the device.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.7.1 — Energy (Part (a))
• Topic 1.7.2 — Work (Part (b))
• Topic 1.7.4 — Power (Part (c))
▶️ Answer/Explanation
(a)
For the correct answer:
chemical (potential energy)
A battery stores energy in the form of chemical potential energy. This energy is released when a chemical reaction occurs within the battery, converting the stored chemical energy into electrical energy to power the device.
(b)
For the correct answer:
63 (J)
(work done =) \(14 \times 4.5\)
(work done =) force × distance (moved in direction of force)
Mechanical work done is calculated using the equation \(W = Fd\). Here, the constant force is \(14\text{ N}\) and the distance moved in the direction of the force is \(4.5\text{ m}\). Multiplying these values gives the total work done: \(14\text{ N} \times 4.5\text{ m} = 63\text{ J}\).
(c)
For the correct answer:
180 (W)
(power input =) 5400 ÷ 30
(power input =) energy input ÷ time OR \(\Delta E ÷ t\)
Power is defined as the rate at which energy is transferred. Using the equation \(P = \frac{\Delta E}{t}\), the total energy input of \(5400\text{ J}\) is divided by the time of \(30\text{ s}\). This calculation yields a power input of \(\frac{5400\text{ J}}{30\text{ s}} = 180\text{ W}\).
