If a $2 k g$ $2 kg$ object moving at $1 \frac{m}{s}$ $1 m/s$ slows to a halt after moving $\frac{1}{4} m$ $1/4 m$ , 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 $1 \frac{m}{s}$ $1 m/s$ slows to a halt after moving $\frac{1}{4} m$ $1/4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-04-12T06:29:26

$0.20$ $0.20$ , rounded to second decimal place.

Explanation:

To find out the retardation due to kinetic friction let us use the equation
$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$ ,

where $v, u, a, s$ $v,u,a,s$ are the final velocity, initial velocity, acceleration and distance moved respectively.
$0^{2} - 1^{2} = 2 \times a \times \frac{1}{4}$ $0^2-1^2=2xxaxx1/4$
or $a = - 2 m s^{- 2}$ $a=-2ms^-2$

Now Force $F = m . a$ $F=m.a$
or Retarding force which is the force $∣ ∣ F_{f} ∣ ∣ = 2 \times 2 = 4 N$ $|F_f|=2xx2=4N$
We know that ths $F_{f} = μ \times Normal Reaction$ $F_f=muxx"Normal Reaction"$ , where $μ$ $mu$ is the coefficient of kinetic friction.

Assuming that the surface is a level surface, $:. "Normal Reaction"=mg$
Taking $g=9.81ms^-2$ , we obtain
$4=muxx2xx9.81$ , solving for $mu$
$mu=4/(2xx9.81)=0.20$ , rounded to second decimal place.

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If a $2 k g$ $2 kg$ object moving at $1 \frac{m}{s}$ $1 m/s$ slows to a halt after moving $\frac{1}{4} m$ $1/4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Explanation:

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If a $2 k g$ $2 kg$ object moving at $1 \frac{m}{s}$ $1 m/s$ slows to a halt after moving $\frac{1}{4} m$ $1/4 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?