Kinetic Energy And Work-energy Theorem
A hockey puck of mass 0.2 kg slides across ice and its speed decreases steadily from 12 m/s to 4 m/s due to friction over a certain distance. What is the magnitude of the work done by friction on the puck?
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About This Question
- Subject
- physics
- Chapter
- work, energy and power
- Topic
- kinetic energy and work-energy theorem
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
12.8 J
According to NCERT Class 11, Chapter 6 (Work, Energy and Power), the work-energy theorem states that the net work done on a body equals the change in its kinetic energy, W = (1/2)m(v² - u²). Here friction is the only horizontal force doing work. The initial kinetic energy is (1/2)(0.2)(12²) = (1/2)(0.2)(144) = 14.4 J. The final kinetic energy is (1/2)(0.2)(4²) = (1/2)(0.2)(16) = 1.6 J. The change in kinetic energy is 1.6 - 14.4 = -12.8 J, so friction does -12.8 J of work; its magnitude is 12.8 J. The option 6.4 J results from forgetting the factor relationship and using only one speed. The option 16 J ignores the final kinetic energy entirely. The option 9.6 J comes from a wrong subtraction of squared speeds. Sign check: friction opposes motion and removes energy, so negative work is expected, and the magnitude 12.8 J is less than the initial 14.4 J, which is physically consistent.
This medium difficulty physics question is from the chapter work, energy and power, covering the topic of kinetic energy and work-energy theorem. It appeared in the 2025 exam.
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