9709/41

Mathematics 9709/41May/June 2017

Cambridge A-Level · Mechanics · worked solutions for every part, with the mark scheme

7
questions
50
marks
75
minutes

Topics Forces and Equilibrium · Newton's Laws of Motion · Kinematics of Motion in a Straight Line · Energy, Work and Power

Q13MEnergy, Work and PowerFree sample

A particle of mass 0.6 kg0.6\text{ kg} is dropped from a height of 8 m8\text{ m} above the ground. The speed of the particle at the instant before hitting the ground is 10 m s110\text{ m s}^{-1}. Find the work done against air resistance.

DifficultyMedium-Easy
Worked solution

Approach

Use the work-energy principle. The loss in gravitational potential energy as the particle falls is the total energy available. This energy is converted into kinetic energy plus the work done against air resistance. Therefore:

work done against air resistance=PE lossKE gain\text{work done against air resistance} = \text{PE loss} - \text{KE gain}

Working

Take g=10 m s2g = 10 \text{ m s}^{-2}.

Loss in gravitational potential energy:

PE loss=mgh=0.6×10×8=48 J\begin{aligned} \text{PE loss} &= mgh \\ &= 0.6 \times 10 \times 8 \\ &= 48 \text{ J} \end{aligned}

Gain in kinetic energy:

KE gain=12mv2=12×0.6×102=30 J\begin{aligned} \text{KE gain} &= \frac{1}{2}mv^2 \\ &= \frac{1}{2} \times 0.6 \times 10^2 \\ &= 30 \text{ J} \end{aligned}

Work done against air resistance:

WD=4830=18 J\text{WD} = 48 - 30 = 18 \text{ J}

Answer

18 J18 \text{ J}
Final answer

18 J

Detailed explanation

Walkthrough

The particle is dropped, so its initial kinetic energy is zero. As it falls, gravitational potential energy is converted into kinetic energy, but some energy is also used to overcome air resistance.

First find the loss in gravitational potential energy:

PE loss=mgh=0.6×10×8=48 J\begin{aligned} \text{PE loss} &= mgh \\ &= 0.6 \times 10 \times 8 \\ &= 48 \text{ J} \end{aligned}

Next find the kinetic energy just before the particle hits the ground:

KE gain=12mv2=12×0.6×102=30 J\begin{aligned} \text{KE gain} &= \frac{1}{2}mv^2 \\ &= \frac{1}{2} \times 0.6 \times 10^2 \\ &= 30 \text{ J} \end{aligned}

If there were no air resistance, these two values would be equal. Here the kinetic energy is smaller than the gravitational potential energy lost, so the missing energy has been transferred to the air as work done against resistance. Therefore:

WD against resistance=4830=18 J\text{WD against resistance} = 48 - 30 = 18 \text{ J}

The work done against air resistance is positive because energy is being removed from the particle's mechanical energy.

Key Takeaways

This question tests the work-energy principle in the presence of a non-conservative force. The total mechanical energy is not conserved because air resistance does negative work on the particle. The loss of gravitational potential energy is split between the gain in kinetic energy and the work done against resistance.

Common Mistakes

  • Forgetting that the particle is dropped, so its initial kinetic energy is zero.
  • Using the final speed as the speed throughout the fall when calculating kinetic energy.
  • Subtracting in the wrong order; the work done against resistance is PE loss minus KE gain.
  • Using g=9.8g = 9.8 when the mark scheme expects g=10g = 10.

Things to Be Careful About

  • Take g=10 m s2g = 10 \text{ m s}^{-2} unless the question states otherwise.
  • The speed given is the speed just before hitting the ground, so it is the final speed used in the kinetic energy calculation.
  • The work done against air resistance is the energy lost from the particle, so it is positive and equal to the difference between the potential energy lost and the kinetic energy gained.
  • Always include units: the final answer is in joules.
Techniques used
calculate gravitational potential energy losscalculate kinetic energy gainapply work-energy principle to find work done against resistance

The rest of this paper

6 more questions
  • Q2Forces and Equilibrium · Newton's Laws of Motion · Kinematics of Motion in a Straight Line6M
  • Q3Forces and Equilibrium · Newton's Laws of Motion6M
  • Q4Energy, Work and Power6M
  • Q5Kinematics of Motion in a Straight Line7M
  • Q6Kinematics of Motion in a Straight Line10M
  • Q7Newton's Laws of Motion · Forces and Equilibrium12M
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