If an object is moving at $5 \frac{m}{s}$ $5 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{3}{g}$ $u_k=3 /g$ , how far will the object continue to move?

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Answer 1 · 2018-05-16T17:26:25

The distance is $= 4.17 m$ $=4.17m$

The coeffcient of kinetic friction is $μ_{k} = \frac{3}{g}$ $mu_k=3/g$

$μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

Let the mass of the object be $= m k g$ $=m kg$

Then,

The normal reaction is $N = m g N$ $N=mg N$

The frictional force is

$F_{r} = μ_{k} N = μ_{k} m g = \frac{3}{g} \cdot m g = 3 m N$ $F_r=mu_kN=mu_kmg=3/g*mg=3mN$

$F = m a$ $F=ma$

The acceleration is $a = \frac{F}{m} = \frac{F_{r}}{m} = - 3 \frac{m}{m} = - 3 m s^{- 2}$ $a=F/m=F_r/m=-3m/m=-3ms^-2$

The initial velocity is $u = 5 m s^{- 1}$ $u=5ms^-1$

Apply the equation of motion

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

The final velocity $v = 0$ $v=0$

$0 = 25 - 2 \times 3 \times s$ $0=25-2xx3xxs$

The distance is

$s = \frac{25}{6} = 4.17 m$ $s=25/6=4.17m$

If an object is moving at 5ms5 m/s over a surface with a kinetic friction coefficient of uk=3gu_k=3 /g, how far will the object continue to move?