If a $2 k g$ $2 kg$ object moving at $80 \frac{m}{s}$ $80 m/s$ slows to a halt after moving $9 m$ $9 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-04-19T08:57:17

The coefficient of kinetic friction is $= 36.28$ $=36.28$

We apply the equation of motion to find the deceleration

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

The initial velocity is $u = 80 m s^{- 1}$ $u=80ms^-1$

The final velocity is $v = 0 m s^{- 1}$ $v=0ms^-1$

The distance is $s = 9 m$ $s=9m$

$0 = 80^{2} + 2 \cdot a \cdot 9$ $0=80^2+2*a*9$

$a = - \frac{80^{2}}{2 \cdot 9}$ $a=-80^2/(2*9)$

$a = - 355.6 m s^{- 2}$ $a=-355.6ms^-2$

The frictional force is

$F_{r} = m a = 2 \cdot 355.6 = 711.1 N$ $F_r=ma=2*355.6=711.1N$

$N = 2 g$ $N=2g$

The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{711.1}{2 g} = 36.28$ $mu_k=F_r/N=711.1/(2g)=36.28$

If a 2kg2 kg object moving at 80ms80 m/s slows to a halt after moving 9m9 m, what is the coefficient of kinetic friction of the surface that the object was moving over?