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Work transfers energy into a spring Motion graphs and constant-acceleration equations Displacement and velocity are vectors From a velocity graph to kinetic energy Free fall through motion and energy Only the force component along the motion does work Resultant work changes kinetic energy Momentum is conserved in more collisions than kinetic energy Photoelectron energy determines its maximum speed Accelerating voltage becomes electron kinetic energy Work, energy and power3.3 Scalar and vector quantities2.3.1 Kinematics3.1.1 Linear motion3.1.2 Work and conservation of energy3.3.1 Kinetic and potential energies3.3.2 Collisions3.5.2 E.m.f. and potential difference4.2.2 The photoelectric effect4.5.2 Elastic potential energy
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Physics connection

Resultant work changes kinetic energy

OCR A AS Physics · Two ideas, one explanation

The total work done by the forces acting on an object equals its change in kinetic energy. Braking is a useful example: resistive forces reduce the kinetic energy.

Wresultant = ½mv² − ½mu²

Follow the reasoning

  1. Calculate initial and final kinetic energies.
  2. Include the work of every force that acts along the displacement.
  3. Relate the net work to the energy change.
Specification outcomes (4)
  • 3.3.1(a) Define work done by a force and the unit joule.
  • 3.3.1(b) Use W = F x cos θ for work done by a force.
  • 3.3.1(c) State and apply the principle of conservation of energy.
  • 3.3.2(a) Define and calculate kinetic energy, Ek = ½mv², and derive it from first principles.
Open the official OCR AS specification ↗

H156 v2.0 · May 2026

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PhysicsUK Solve15 marks

A technician calibrates the spring bumper of a small instrument trolley. Figure 1 shows the force F needed to compress the spring by x from its natural length. The graph is a straight line through (0, 0), (0.020 m, 1.2 N), (0.040 m, 2.4 N) and (0.060 m, 3.6 N). The trolley has mass 0.60 kg. It is released from rest against the spring at compression 0.060 m on a horizontal track. A constant resistive force of 0.60 N opposes the motion as soon as the trolley moves. The spring's mass is negligible and its deformation is elastic. The trolley loses contact with the spring at its natural length. There are no other energy losses. (a) Use the gradient of the force-compression graph to determine the spring constant. [2] (b) Use the graph to calculate the work needed to compress the spring from its natural length to 0.060 m. [2] (c) Calculate the acceleration just after release, stating its direction relative to the bumper. [3] (d) Calculate the speed when the trolley loses contact with the spring. Account for the work done against resistance. [3] (e) Explain why a single constant-acceleration equation cannot describe the whole interval while the trolley is in contact with the spring. [2] (f) Calculate the distance travelled after losing contact before the trolley stops. Use a constant-acceleration equation for this part. [3]

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