If a $6 k g$ $6 kg$ object moving at $1 \frac{m}{s}$ $1 m/s$ slows down to a halt after moving $3 m$ $3 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-12-29T08:36:53

$μ = \frac{1}{80}$ $mu=1/(80)$

We know that,
$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$\frac{1}{4} = 1 + 2 \times 3 \times a$ $1/4 = 1 +2xx3xxa$

$- \frac{3}{4} = 6 \times a$ $-3/4 = 6xxa$

$a = - \frac{1}{8}$ $a= -1/8$

By Newton's law,

$F = m a$ $F=ma$

$f = 6 \times \frac{1}{8}$ $f= 6xx1/8$

$μ N = \frac{3}{4}$ $mu N=3/4$

$μ = \frac{3}{4 \times 60}$ $mu=3/(4xx60)$

$μ = \frac{1}{80}$ $mu=1/(80)$

If a 6kg6 kg object moving at 1ms1 m/s slows down to a halt after moving 3m3 m, what is the coefficient of kinetic friction of the surface that the object was moving over?