If a $\frac{1}{24} k g$ $1/24 kg$ object moving at $\frac{1}{3} \frac{m}{s}$ $1/3 m/s$ slows to a halt after moving $\frac{10}{27} m$ $10/27 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-04-15T08:05:55

The coefficient of kinetic friction is $= 0.015$ $=0.015$

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The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

$F_{f} = m a$ $F_f=ma$

We apply the equation of motion

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$0 = {(\frac{1}{3})}^{2} + 2 \cdot a \cdot \frac{10}{27}$ $0=(1/3)^2+2*a*10/27$

$\frac{20}{27} a = - \frac{1}{9}$ $20/27a=-1/9$

$a = - \frac{27}{9} \cdot \frac{1}{20} = - \frac{3}{20} m s^{- 2}$ $a=-27/9*1/20=-3/20ms^-2$

The negative sign indicates that it is a deceleration

Therefore,

$F_{r} = \frac{1}{24} \cdot - \frac{3}{20} = - \frac{1}{160} N$ $F_r=1/24*-3/20=-1/160N$

Therefore,

$μ_{k} = \frac{1}{160} \cdot \frac{1}{\frac{1}{24} g}$ $mu_k=1/160*1/(1/24g)$

$= \frac{1}{65.3} = 0.015$ $=1/65.3=0.015$

If a 124kg1/24 kg object moving at 13ms1/3 m/s slows to a halt after moving 1027m10/27 m, what is the coefficient of kinetic friction of the surface that the object was moving over?