A 2\text{kg} box slides 5\text{m} across a floor with \muk = 0.30 and starts at 4.0\text{m/s}. Using g \approx 10\text{m/s}^2, its final speed is

0 Friction work: Wf = \muk m g d = (0.30)(2)(10)(5) = 30\text{J}. Initial KE: \tfrac{1}{2}(2)(16) = 16\text{J}. Since Wf K0, the box stops before 5\text{m}.

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Acemy

Conservation of Energy

AP Physics 1· difficulty 3/5

2~\text{kg}

A
$3.0~\text{m/s}$
B
$2.0~\text{m/s}$
C
$0$
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D
$1.0~\text{m/s}$

Explanation

W_f = \mu_k m g d = (0.30)(2)(10)(5) = 30~\text{J}

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