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9702 · 5.2

Gravitational potential energy and kinetic energy — practice questions

Practice and worked examples for 9702 Gravitational potential energy and kinetic energy. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A 2.0 kg ball is dropped from a height of 15 m above the ground. Calculate its speed just before it hits the ground, assuming negligible air resistance. (Take g=9.81g = 9.81 m s\textsuperscript{-2})

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  1. Principle: By the principle of conservation of energy, the initial GPE is converted into final KE. GPEinitial=KEfinalGPE_{initial} = KE_{final}
  2. Formulate: Write the equation using the formulas for GPE and KE. mgh=12mv2mgh = \frac{1}{2}mv^2
  3. Simplify and Solve: The mass mm cancels out from both sides. gh=12v2gh = \frac{1}{2}v^2 v2=2ghv^2 = 2gh v=2ghv = \sqrt{2gh}
  4. Substitute Values: v=2×(9.81 m s−2)×(15 m)v = \sqrt{2 \times (9.81 \text{ m s}^{-2}) \times (15 \text{ m})} v=294.3≈17.155 m s−1v = \sqrt{294.3} \approx 17.155 \text{ m s}^{-1}
  5. Final Answer: The speed of the ball just before it hits the ground is approximately 17.2 m s−117.2 \text{ m s}^{-1} (to 3 significant figures).

Worked example 2

A box of mass 5.0 kg is pushed up a rough slope inclined at 30° to the horizontal. It is given an initial speed of 8.0 m/s at the bottom and travels 4.0 m up the slope before coming to rest. Calculate the work done against the resistive forces. (Take g=9.81g = 9.81 m s\textsuperscript{-2})

Show solution outline
  1. Initial Energy: Calculate the total mechanical energy at the bottom of the slope. We define the initial height as h=0h=0. KEinitial=12mv2=12×(5.0 kg)×(8.0 m s−1)2=160 JKE_{initial} = \frac{1}{2}mv^2 = \frac{1}{2} \times (5.0 \text{ kg}) \times (8.0 \text{ m s}^{-1})^2 = 160 \text{ J} GPEinitial=mgh=0 JGPE_{initial} = mgh = 0 \text{ J} Einitial=KEinitial+GPEinitial=160 JE_{initial} = KE_{initial} + GPE_{initial} = 160 \text{ J}
  2. Final Energy: Calculate the total mechanical energy after it travels 4.0 m up the slope and comes to rest (v=0v=0). First, find the vertical height gained: hfinal=dsin⁡(θ)=(4.0 m)×sin⁡(30°)=2.0 mh_{final} = d \sin(\theta) = (4.0 \text{ m}) \times \sin(30°) = 2.0 \text{ m}. KEfinal=0 JKE_{final} = 0 \text{ J} (since it comes to rest) GPEfinal=mghfinal=(5.0 kg)×(9.81 m s−2)×(2.0 m)=98.1 JGPE_{final} = mgh_{final} = (5.0 \text{ kg}) \times (9.81 \text{ m s}^{-2}) \times (2.0 \text{ m}) = 98.1 \text{ J} Efinal=KEfinal+GPEfinal=98.1 JE_{final} = KE_{final} + GPE_{final} = 98.1 \text{ J}
  3. Work Done Against Resistance: The work done by resistive forces is the energy lost from the mechanical system. Wresistive=Einitial−EfinalW_{resistive} = E_{initial} - E_{final} Wresistive=160 J−98.1 J=61.9 JW_{resistive} = 160 \text{ J} - 98.1 \text{ J} = 61.9 \text{ J}
  4. Final Answer: The work done against resistive forces is 61.9 J61.9 \text{ J}.
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