If a $5 k g$ $5 kg$ object moving at $6 \frac{m}{s}$ $6 m/s$ slows to a halt after moving $2 m$ $2 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2018-01-25T09:09:14

The coefficient of kinetic friction is $= 0.92$ $=0.92$

The mass of the object is $m = 5 k g$ $m=5kg$

The initial speed of the object is $u = 6 m s^{- 1}$ $u=6ms^-1$

The final speed of the object is $v = 0 m s^{- 1}$ $v=0ms^-1$

The distance is $s = 2 m$ $s=2m$

Apply the equation of motion

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

to calculate the acceleration

The acceleration is

$a = \frac{v^{2} - u^{2}}{2 s} = \frac{0^{2} - 6^{2}}{2 \cdot 2} = - \frac{36}{4} = - 9 m s^{- 2}$ $a=(v^2-u^2)/(2s)=(0^2-6^2)/(2*2)=-36/4=-9ms^-2$

$F = m a$ $F=ma$

The frictional force is $F_{r} = 5 \cdot 9 = 45 N$ $F_r=5*9=45N$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal force is $N = m g = 5 \cdot 9.8 = 49 N$ $N=mg=5*9.8=49N$

The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{45}{49} = 0.92$ $mu_k=F_r/N=45/49=0.92$

If a 5kg5 kg object moving at 6ms6 m/s slows to a halt after moving 2m2 m, what is the coefficient of kinetic friction of the surface that the object was moving over?