A 2.0 kg block is released from rest at the top of a frictionless incline that is 4.0 m high. At the bottom of the incline, the block slides onto a rough horizontal surface with a coefficient of kinetic friction of 0.30. 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
A. 13.3 m
Principle or equation
Conservation of mechanical energy on the frictionless incline gives the speed at the bottom: mgh = (1/2)mv². On the rough surface, the work done by friction equals the change in kinetic energy: μmg d = (1/2)mv².
Why this answer is correct
At the bottom, v² = 2gh = 2(10)(4.0) = 80 m²/s². On the rough surface, friction force = μmg = 0.30 × 2.0 × 10 = 6.0 N. Work done by friction = 6.0 × d. Setting equal to kinetic energy (1/2)(2.0)(80) = 80 J gives 6.0 d = 80, so d = 13.33 m ≈ 13.3 m.
Example
If the height were 2 m and μ = 0.2, then v² = 40, KE = 40 J, friction = 4 N, so d = 10 m.
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