If a $\frac{1}{2} k g$ $1/2 kg$ object moving at $\frac{7}{4} \frac{m}{s}$ $7/4 m/s$ slows to a halt after moving $\frac{3}{8} m$ $3/8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2017-01-21T10:25:20

The answer is $μ_{k} = 1.25$ $mu_k=1.25$

We solve in the horizontal direction $\to^{+}$ $rarr^+$

$u = \frac{7}{4} m s^{- 2}$ $u=7/4ms^-2$

$v = 0$ $v=0$

$s = \frac{3}{8} m$ $s=3/8m$

We use the equation

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

$0 = \frac{49}{16} + 2 \cdot \frac{3}{8} \cdot a$ $0=49/16+2*3/8*a$

$a = - \frac{49}{16} \cdot \frac{4}{3} = - \frac{49}{12} m s^{- 2}$ $a=-49/16*4/3=-49/12ms^-2$

By Newton`s second Law

$F_{r} = m a = - \frac{49}{4} \cdot \frac{1}{2} = - \frac{49}{8} N$ $F_r=ma=-49/4*1/2=-49/8N$

The reaction is $N = \frac{1}{2} g = \frac{1}{2} \cdot 9.8 = 4.9 N$ $N=1/2g=1/2*9.8=4.9N$

The coefficient of kinetic friction is $μ_{k} = \frac{F_{r}}{N} = \frac{49}{8} \cdot \frac{1}{4.9} = 1.25$ $mu_k=F_r/N=49/8*1/4.9=1.25$

If a 12kg1/2 kg object moving at 74ms7/4 m/s slows to a halt after moving 38m3/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?