Work-energy Via Newton’s Laws
Easyphysics
A 2 kg block is pushed across a rough horizontal surface by a 16 N force. It moves at constant velocity. What is the coefficient of kinetic friction? (g = 10 m/s²)
Select the correct option:
Solution
Incorrect! Answer:
0.80
- Equilibrium Condition: 'Constant velocity' means zero acceleration (a=0). By Newton's First Law, net force must be zero.
- Force Balance:
- Applied Force (F) = Kinetic Friction (fk)
- 16 N =μkN.
- Normal Force: On a horizontal surface, N=mg=2×10=20 N.
- Solve for Coefficient:
- 16=μk×20
- μk=2016=108=0.80.
- Result: The coefficient of kinetic friction is 0.80.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- work-energy via newton’s laws
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
0.80
- Equilibrium Condition: 'Constant velocity' means zero acceleration (a=0). By Newton's First Law, net force must be zero.
- Force Balance:
- Applied Force (F) = Kinetic Friction (fk)
- 16 N =μkN.
- Normal Force: On a horizontal surface, N=mg=2×10=20 N.
- Solve for Coefficient:
- 16=μk×20
- μk=2016=108=0.80.
- Result: The coefficient of kinetic friction is 0.80.
This easy difficulty physics question is from the chapter laws of motion, covering the topic of work-energy via newton’s laws. It appeared in the 2025 exam.
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