NCEA Level 1 Physics 92047 Energy in physical systems: past papers, practice and marking
- NCEA Level 1
- Physics
- AS92047
- 5 credits
- External exam
About 92047
AS92047, Demonstrate understanding of a physical system using energy concepts. Falling and lifted objects, work and power, specific and latent heat, heat transfer, circuits and electrical power: the 5-credit Level 1 external, in the six question types it keeps reusing.
What 92047 covers
- Energy changesEnergy transformations in falling, lifted and swinging objects, Ep = mgh, Ek = ½mv², speed from conservation of energy, stating the assumption, energy lost to thermal energy through air resistance
- Work and powerW = Fd, when work is and is not done, work linked to changes in Ek and Ep, an average resistive force from lost energy, power as the rate of transferring energy, P = E/t, times in seconds
- HeatTemperature against thermal energy, Ethermal = mcΔT with the temperature change, units J kg⁻¹ °C⁻¹, latent heat of fusion and Ethermal = mL, the flat part of a heating graph, measured values against accepted values
- Heat transferConduction, convection and radiation in a context, labelled convection current diagrams with particle motion and density, trapped air, vacuum flasks and insulated homes, the average rate of thermal energy gain
- CircuitsSeries and parallel circuits, ammeters in series and voltmeters in parallel, V = IR from data or a graph, a variable resistor sharing a fixed supply voltage, the filament lamp
- Electrical powerP = VI and E = Pt in heaters, jugs and appliances, hours and minutes in seconds, how a bigger current gives more thermal energy each second, why a measured value differs from a calculated one
92047 past papers
Open a 92047 past paper beside your own Ako page, then get it marked, part by part.
Work through a past paper All NCEA Level 1 Physics past papers
Practice questions
Question 1: Energy changes
In netball shooting practice, a player throws a ball of mass 0.40 kg straight up. It leaves her hands at 8.0 m s⁻¹. Use g = 10 N kg⁻¹.
- Describe the energy changes of the ball on the way up and on the way back down.(up to Achieved)
- Calculate the kinetic energy of the ball as it leaves her hands.(up to Achieved)
Check your answer
Part a: what gets the marks
- On the way up Ek changes into Ep, on the way down Ep changes back into Ek, and each way some energy becomes thermal energy.
A full answer: On the way up, the ball slows down as kinetic energy is transformed into gravitational potential energy. At the top it is momentarily still. On the way down, Ep is transformed back into Ek. In both directions a little energy is transferred to thermal energy by air resistance.
Part b: what gets the marks
- Ek = ½mv²: Ek = ½ × 0.40 × 8.0² = 12.8 J
A full answer: Ek = ½mv² = ½ × 0.40 × 8.0² = 12.8 J.
What gets each grade
Question 2: Energy changes
On a building site, an electric hoist lifts a bundle of roofing iron of mass 60 kg from the ground to a roof 12 m high. The hoist's motor runs at a power of 600 W and the lift takes 20 s, at a steady speed. Use g = 10 N kg⁻¹.
- Describe the energy changes as the hoist lifts the roofing iron.(up to Achieved)
- Calculate the electrical energy supplied to the motor during the lift.(up to Achieved)
Check your answer
Part a: what gets the marks
- As the hoist lifts the load, electrical energy in the motor becomes gravitational potential energy of the roofing iron, plus thermal energy and sound in the motor.
A full answer: Electrical energy supplied to the motor is transformed into gravitational potential energy of the roofing iron as it rises (and a small amount of kinetic energy while it moves), with some transferred to thermal energy and sound in the motor and hoist.
Part b: what gets the marks
- E = Pt: E = Pt = 600 × 20 = 12 000 J
A full answer: E = Pt = 600 × 20 = 12 000 J.
What gets each grade
Question 3: Energy changes
On a building site, an electric hoist lifts a bundle of roofing iron of mass 60 kg from the ground to a roof 12 m high. The hoist's motor runs at a power of 600 W and the lift takes 20 s, at a steady speed. Use g = 10 N kg⁻¹.
- Calculate the gravitational potential energy gained by the roofing iron.(up to Achieved)
- The motor is supplied with 12 000 J of electrical energy during the lift. Explain why this is different from the gravitational potential energy the roofing iron gains, and where the difference has gone.(up to Excellence)
Check your answer
Part a: what gets the marks
- Ep = mgh: Ep = 60 × 10 × 12 = 7200 J
A full answer: Ep = mgh = 60 × 10 × 12 = 7200 J.
Part b: what gets the marks
- The difference is 12 000 J − 7200 J = 4800 J.
- At a steady speed almost none of the energy is kinetic energy (only about 11 J).
- The 4800 J is transferred to thermal energy and sound by friction in the motor, gears and pulley and by the current heating the motor's coils, and spreads into the hoist and the air.
A full answer: The motor is supplied with 12 000 J but the roofing iron gains only 7200 J of Ep, so 4800 J has not gone into the load. Very little of it is kinetic energy: at a steady 0.6 m s⁻¹ the load has only about 11 J of Ek. The rest is transferred to thermal energy and sound: friction between the moving parts of the motor, gears and pulley heats them, and the current in the motor's wires heats them too. This thermal energy spreads into the hoist and the air. Energy is conserved: 12 000 J in = 7200 J Ep + about 4800 J of thermal energy and sound (and a tiny Ek).