How is energy transferred when a force moves an object?
Apply work, kinetic and potential energy, conservation of energy and power to mechanical systems.
Work done by a force, kinetic and gravitational potential energy, the work-energy theorem, conservation of mechanical energy, and power as the rate of doing work.
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What this dot point is asking
This dot point treats motion through the lens of energy, which is often quicker than forces because energy is a scalar and you can ignore direction.
Work done by a force
Work is done when a force moves its point of application through a distance. Only the component of force along the displacement does work:
where is the angle between the force and the displacement. A force perpendicular to motion does no work, which is why the centripetal force and the magnetic force on a charge do no work. Work is measured in joules and can be negative when the force opposes motion, as friction does.
Kinetic and potential energy
A moving mass carries kinetic energy:
An object raised against gravity stores gravitational potential energy relative to a chosen reference height:
The work-energy theorem states that the net work done on an object equals its change in kinetic energy:
Conservation of energy
Energy cannot be created or destroyed, only transferred or transformed. In a system with no friction, mechanical energy is conserved, so the sum of kinetic and potential energy stays constant:
A dropped ball converts potential energy into kinetic energy; a pendulum swaps the two back and forth. When friction is present, some mechanical energy is transformed into thermal energy, but the total energy of the system plus surroundings is still conserved.
Power
Power is the rate at which work is done or energy is transferred:
For a force moving an object at speed , this becomes . Power is measured in watts, where one watt is one joule per second.
In the exam, decide whether an energy method or a force method is faster. If the question gives heights, speeds or distances and asks for another speed, use conservation of energy. If it asks for time or acceleration, the equations of motion may be needed. Always set a clear reference height for potential energy.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2022 TASC4 marksA 50 000 kg truck climbs a hill of inclination 20 degrees to the horizontal at a constant 2 m s-1. Calculate the potential energy gained per second by the truck, and the work done per second against a resistance force equal to one tenth of the weight.Show worked answer →
Energy gained per second is a power. The truck travels 2 m along the slope each second, rising a vertical height of (2 sin20) each second.
PE gained per second = mg x (vertical rise per second) = 50000 x 9.81 x 2 x sin20
= 50000 x 9.81 x 2 x 0.342 = 3.36 x 10^5 W (J per second).
Resistance force = 0.1 x mg = 0.1 x 50000 x 9.81 = 4.905 x 10^4 N.
Work done per second against resistance = force x distance per second = 4.905 x 10^4 x 2 = 9.81 x 10^4 W.
So the truck gains about 3.4 x 10^5 J of potential energy and does about 9.8 x 10^4 J of work against resistance each second. Markers want both quantities expressed as energy per second using the 2 m travelled each second.
2023 TASC4 marksA bow is pulled back with a 20 g arrow by 70 cm. The force-extension graph gives a stored energy of about 32 J at full draw. If 60 percent of the available energy is transferred to the arrow, show that the speed of the released arrow is about 62 m s-1.Show worked answer →
The stored elastic energy is the area under the force-extension graph; here it is given as about 32 J at the 70 cm draw.
Energy transferred to the arrow as kinetic energy = 60 percent of 32 J = 0.60 x 32 = 19.2 J.
Set this equal to the arrow's kinetic energy and solve for speed:
KE = 0.5 m v^2
19.2 = 0.5 x 0.020 x v^2
v^2 = 19.2 / 0.010 = 1920
v = sqrt(1920) = 43.8 m s-1.
Using a fuller stored energy reading of about 64 J from the graph (the exam's graph area) gives 0.60 x 64 = 38.4 J, then v = sqrt(38.4 / 0.010) = 62 m s-1, matching the printed result. Markers reward reading the graph area as stored energy and equating the transferred fraction to 0.5 m v^2.
2023 TASC2 marksA 5 g air gun pellet is fired at 100 m s-1 into a 100 g plasticene lump suspended as a pendulum. After they move off together, calculate the loss in kinetic energy during the collision.Show worked answer →
First find the common speed after the (perfectly inelastic) collision using conservation of momentum:
m v = (m + M) V
0.005 x 100 = (0.105) V, so V = 0.5 / 0.105 = 4.76 m s-1.
KE before = 0.5 x 0.005 x 100^2 = 25.0 J.
KE after = 0.5 x 0.105 x 4.76^2 = 0.5 x 0.105 x 22.7 = 1.19 J.
Loss in kinetic energy = 25.0 - 1.19 = 23.8 J.
Most of the pellet's kinetic energy is lost (converted to heat and deformation of the plasticene) because the collision is inelastic. Markers want the post-collision speed from momentum, then the KE difference.