A 2.0 kg block is released from rest at the top of a frictionless incline that is 3.0 m high. At the bottom of the incline, the block slides onto a rough horizontal surface with a coefficient of kinetic friction of 0.40. 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. 7.5 m
Principle or equation
Mechanical energy is conserved on the frictionless incline, so the kinetic energy at the bottom equals mgh. On the rough surface, the work done by friction (f_k d = μmgd) removes this kinetic energy.
Why this answer is correct
At the bottom, KE = mgh = (2.0)(10)(3.0) = 60 J. On the rough surface, friction does work W = -μmgd = -(0.40)(2.0)(10)d = -8.0d. Setting KE + W = 0 gives 60 - 8.0d = 0, so d = 7.5 m.
Example
If a 1 kg block slides down a 2 m high incline and then slides on a surface with μ = 0.2, then KE = 20 J, friction work = -2.0d, so d = 10 m.
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