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Springs and Newton’s second law Forces explain changes in motion Motion3.1 Forces in action3.2 Dynamics3.2.1 Springs3.4.1
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Dynamics

OCR A AS Physics · 3.2.1

Explore dynamics within forces in action. Start with the object and draw all forces acting on it. Their vector sum determines acceleration; zero resultant force gives constant velocity, including rest.

Specification outcomes (6)
  • 3.2.1(a) Apply Newton’s second law, F = ma.
  • 3.2.1(b) Define the newton as the unit of force.
  • 3.2.1(c) Understand weight of an object, W = mg.
  • 3.2.1(d) Define and identify tension, normal contact force, upthrust and friction.
  • 3.2.1(e) Draw and interpret free-body diagrams.
  • 3.2.1(f) Describe one- and two-dimensional motion under constant force.
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H156 v2.0 · May 2026

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PhysicsUK Solve13 marks

An athlete and racing wheelchair have a combined mass of $84.0\,\mathrm{kg}$. At the start of a straight training run, the wheelchair is already moving at $2.00\,\mathrm{m\,s^{-1}}$. For the next $1.50\,\mathrm{s}$, the athlete produces a constant forward driving force of $420\,\mathrm{N}$. A constant rolling resistance of $84.0\,\mathrm{N}$ acts backward. After this $1.50\,\mathrm{s}$ push, the athlete stops producing a driving force. Assume the rolling resistance remains $84.0\,\mathrm{N}$ until the wheelchair stops. (a) Calculate the resultant force and acceleration during the push. [3] (b) Calculate: (i) the speed at the end of the push; (ii) the distance travelled during the push. [4] (c) Calculate the acceleration after the driving force is removed. [2] (d) Calculate: (i) the time taken to stop after the push; (ii) the distance travelled after the push. [3] (e) Calculate the total distance travelled from the start of the stated motion until the wheelchair stops. [1]

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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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