If a $2 k g$ $2 kg$ object moving at $\frac{1}{2} \frac{m}{s}$ $1/2 m/s$ slows to a halt after moving $5 m$ $5 m$ , 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 $\frac{1}{2} \frac{m}{s}$ $1/2 m/s$ slows to a halt after moving $5 m$ $5 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2018-03-13T07:33:18

$μ_{k} = 0.002548$ $mu_k = 0.002548$

Explanation:

Well, the force of friction $F_{f}$ $F_f$ is defined as the following...

$F_{f} = μ_{k} N$ $F_f = mu_k N$

And the normal is given by...

$N = m g$ $N = mg$

$N = (2) (9.81)$ $N = (2)(9.81)$

$N = 19.62 N$ $N = 19.62N$

So now, we need to calculate an acceleration to get the force of friction. We are assuming in this question that $F_{d r i v e} = F_{f}$ $F_(drive) = F_f$ because all the energy that the mass possess is lost to friction.

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

$0^{2} = {0.5}^{2} - 2 (5) a$ $0^2 = 0.5^2 -2(5)a$

$0 = 0.25 - 10 a$ $0 = 0.25-10a$

$a = \frac{- 0.25}{10}$ $a = (-0.25)/(10)$

$a = - 0.025 m s^{- 2}$ $a= -0.025 ms^-2$

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

$F_{f} = (1) | - 0.025 |$ $F_f=(1)abs(-0.025)$

$F_{f} = 0.025 N$ $F_f=0.025N$

$μ_{k} = \frac{F_{f}}{N}$ $mu_k = F_f/N$

$μ_{k} = \frac{0.025}{9.81}$ $mu_k = 0.025/9.81$

$μ_{k} = 0.002548$ $mu_k = 0.002548$

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If a $2 k g$ $2 kg$ object moving at $\frac{1}{2} \frac{m}{s}$ $1/2 m/s$ slows to a halt after moving $5 m$ $5 m$ , 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 $\frac{1}{2} \frac{m}{s}$ $1/2 m/s$ slows to a halt after moving $5 m$ $5 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?