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# dynamics physics?

A curling stone with mass 20.0 kg leaves the curler’s hand at a speed of 0.885 m/s. It slides 31.5 m down the rink before coming to rest. What is the average force of friction?

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- billrussell42Lv 73 months agoFavorite Answer
a = v²/2d = 0.885²/2•31.5 = 0.0124 m/s²

F = ma = 20•0.0124 = 0.249 N

- Wayne DeguManLv 73 months ago
Considering energy principles we can say:

K.E loss = work done by friction force

so, (1/2)(20)(0.885)² = Friction x 31.5

Hence, Friction = 10(0.885)²/31.5 = 0.249 Newtons

:)>

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