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

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Answer 1 · 2017-05-19T06:30:24

The coefficient of friction is $= 0.051$ $=0.051$

The coefficient of kinetic friction is $μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

The normal force is $N = 1 g N$ $N=1gN$

Initial velocity is $u = 1 m s^{- 1}$ $u=1ms^-1$

Final velocity is $v = 0 m s^{- 1}$ $v=0ms^-1$

Distance is $s = 1 m$ $s=1m$

We apply the equation of motion

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

$0 = 1 + 2 a$ $0=1+2a$

$a = - \frac{1}{2} m s^{- 2}$ $a=-1/2ms^-2$

The negative sign indicates that it is a deceleration.

$F = m a$ $F=ma$

$F_{r} = 1 \cdot \frac{1}{2} = \frac{1}{2} N$ $F_r=1*1/2=1/2N$

Therefore,

$μ_{k} = \frac{\frac{1}{2}}{1 g} = \frac{1}{2 g} = 0.051$ $mu_k=(1/2)/(1g)=1/(2g)=0.051$

If a 1 kg1kg1 kg object moving at 1 m/s1ms1 m/s slows down to a halt after moving 1 m1m1 m, what is the friction coefficient of the surface that the object was moving over?