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A 2 kg marble block rests on ice. If it is set moving with a speed of 6 m/s and comes to rest due to friction in 10 s, what is the coefficient of friction?

1. 0.02
2. 0.03
3. 0.04
4. 0.06

Correct answer: 0.06

Solution

Deceleration a = v/t = 6/10 = 0.6 m/s^2. Friction provides this: mu = a/g = 0.6/10 = 0.06 (about 0.06 even with g = 9.8).

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