A 2.0 kg block is released from rest at the top of a frictionless incline that is 5.0 m high. At the bottom of the incline, the block slides onto a rough horizontal surface with a coefficient of kinetic friction of 0.20. How far does the block slide on the rough surface before coming to rest? (Take g = 10 m/s²)
Show the answer and explanation
Correct answer
C. 25 m
Principle or equation
Conservation of energy: potential energy at top converts to kinetic energy at bottom, then work done by friction dissipates it. mgh = μmgd, so d = h/μ.
Why this answer is correct
At the bottom, kinetic energy = mgh = 2.0 × 10 × 5.0 = 100 J. On the rough surface, friction force = μmg = 0.20 × 2.0 × 10 = 4.0 N. Work done by friction = friction force × distance = 4.0 × d. Set equal to 100 J: d = 100 / 4.0 = 25 m.
Example
If h = 2 m and μ = 0.5, d = 2/0.5 = 4 m.
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