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A Levels Physics (9702)•9702/14/M/J/25
Question 14 from 9702/14/M/J/25

Explanation

Work requires net force parallel to displacement

Steps:

  • Define work as W=F⃗⋅d⃗=Fdcos⁡θW = \vec{F} \cdot \vec{d} = F d \cos\thetaW=F⋅d=Fdcosθ, where positive work occurs if force and displacement align.
  • Evaluate each option for applied force and object motion.
  • Identify scenarios with zero displacement or zero net force component along motion.
  • Select the case with clear force and parallel displacement.

Why C is correct:

  • Pushing applies force with a component parallel to the ramp's displacement, so W=F∥d>0W = F_\parallel d > 0W=F∥​d>0 by work-energy theorem.

Why the others are wrong:

  • A: Constant velocity on frictionless surface means net force is zero, so no work done.
  • B: Object is held stationary, so displacement d=0d = 0d=0, work is zero.
  • D: Object is stationary while floating, so displacement d=0d = 0d=0, work is zero.

Final answer: C

Topic: Gravitational potential energy and kinetic energy

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