If a $\frac{3}{2} k g$ $3/2 kg$ object moving at $\frac{3}{4}$ $3/4$ $m s^{- 1}$ $ms^-1$ slows to a halt after moving $\frac{5}{8} m$ $5/8 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2016-08-01T21:27:59

First find the acceleration, then use it to find the magnitude of the frictional force, and use that to find the coefficient of friction.

First find the acceleration:

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

Rearranging to make $a$ $a$ the subject:

$a = \frac{v^{2} - u^{2}}{2 s}$ $a = (v^2-u^2)/(2s)$

$a = \frac{0^{2} - {(\frac{3}{4})}^{2}}{2 \times (\frac{5}{8})} = - 0.45$ $a = (0^2-(3/4)^2)/(2xx(5/8)) = -0.45$ $m s^{- 2}$ $ms^-2$

The negative sign just shows that the object is decelerating - its acceleration is in the opposite direction to the initial velocity.

Now to find the frictional force:

$F = m a = (\frac{3}{2}) \times (0.45) = 0.675$ $F = ma = (3/2)xx(0.45) = 0.675$ $N$ $N$

The frictional force is given by $F_{frict} = μ F_{norm}$ $F_"frict" = muF_"norm"$ , where the normal force is $F_{norm} = m g$ $F_"norm"=mg$ .

Rearranging to make the frictional coefficient the subject:

$μ = \frac{F_{frict}}{F_{norm}} = \frac{0.675}{\frac{3}{2} \times 9.8} = 0.022$ $mu=(F_"frict")/(F_"norm") = 0.675/(3/2xx9.8) = 0.022$

(frictional coefficients have no units, since they are ratios of forces)

(note: technically, a coefficient of friction is between two bodies, not a property of the surface alone)

If a 32kg3/2 kg object moving at 343/4 ms−1ms^-1 slows to a halt after moving 58m5/8 m, what is the coefficient of kinetic friction of the surface that the object was moving over?