If a 8 kg object moving at 16 m/s slows to a halt after moving 80 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

Jan 20, 2018

0.16

Explanation:

Clearly,here the frictional force caused retardation of the moving object and brought it to rest.

So,using ${v}^{2} = {u}^{2} - 2 a s$ (all symbols are bearing their conventional meaning)(given, $v = 0$ ,$u = 16 \frac{m}{s}$ and $s = 80$)

We get, $a = 1.6 \frac{m}{s} ^ 2$

Now,if the coefficient of kinetic friction is $\mu$ , frictional force($f$) acting on the body was $\mu \cdot N$ or $\mu \cdot m g$ i.e $\mu \cdot 80$

So,we using $f = m a$ (where, $m = 8 K g$ and $a = 1.6 \frac{m}{s} ^ 2$)

We, get $\mu \cdot 80 = 12.8$
Or, $\mu = 0.16$

Jan 20, 2018

The coefficient of kinetic friction is $= 0.16$

Explanation:

The mass of the object is $m = 8 k g$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The Normal reaction is

$N = m g = 8 \times 9.8 = 78.4 N$

The initial velocity is $u = 16 m {s}^{-} 1$

The final velocity is $v = 0 m {s}^{-} 1$

The distance is $s = 80 m$

Apply the equation of motion,

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

The acceleration is

$a = \frac{{v}^{2} - {u}^{2}}{2 s} = \frac{{0}^{2} - {16}^{2}}{2 \cdot 80} = - 1.6 m {s}^{-} 2$

According to Newton's Second Law

$F = m a$

The force of friction is

${F}_{r} = 8 \cdot 1.6 = 12.8 N$

The coefficient of kinetic friction is

${\mu}_{k} = {F}_{r} / N = \frac{12.8}{78.4} = 0.16$