If a $2 k g$ $2 kg$ object moving at $80 \frac{m}{s}$ $80 m/s$ slows to a halt after moving $4 m$ $4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $2 k g$ $2 kg$ object moving at $80 \frac{m}{s}$ $80 m/s$ slows to a halt after moving $4 m$ $4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-11-05T06:22:22

The coefficient of kinetic friction is $= 81.6$ $=81.6$

Explanation:

Apply the equation of motion

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

to determine the acceleration $a$ $a$

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

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

The distance is $s = 4 m$ $s=4m$

The acceleration is

$a = \frac{v^{2} - u^{2}}{2 s} = \frac{0^{2} - 80^{2}}{2 \times 4} = - 80 \cdot \frac{80}{8} = - 800 m s^{- 2}$ $a=(v^2-u^2)/(2s)=(0^2-80^2)/(2 xx 4)=-80*80/8=-800ms^-2$

The frictional force is $F_{r} = m a = 2 \times 800 = 1600 N$ $F_r=ma=2xx800=1600N$

Let the acceleration due to gravity be $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal force is $N = m g = 2 \cdot 9.8 = 19.6 N$ $N=mg=2*9.8=19.6N$

The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{1600}{19.6} = 81.6$ $mu _k=F_r/N=1600/19.6=81.6$

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If a $2 k g$ $2 kg$ object moving at $80 \frac{m}{s}$ $80 m/s$ slows to a halt after moving $4 m$ $4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer

Explanation:

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Science

Math

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If a 2kg2 kg object moving at 80ms80 m/s slows to a halt after moving 4m4 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

1 Answer

Explanation:

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If a $2 k g$ $2 kg$ object moving at $80 \frac{m}{s}$ $80 m/s$ slows to a halt after moving $4 m$ $4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?