If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ 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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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ 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 · 2018-01-26T10:48:34

The distance is $= 2.67 m$ $=2.67m$

Explanation:

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

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

The initial velocity is $u = 4 m s^{- 1}$ $u=4ms^-1$

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

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

The normal force is $N = m g N$ $N=mgN$

The frictional force is $= F_{r} N$ $=F_r N$

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

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

According to Newton's Second Law

$F = m a$ $F=ma$

$F_{r} = - m a$ $F_r=-ma$

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

Apply the equation of motion,

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

The distance is

$s = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - 4^{2}}{2 \cdot (- 3)} = \frac{16}{6} = \frac{8}{3} = 2.67 m$ $s=(v^2-u^2)/(2a)=(0-4^2)/(2*(-3))=16/6=8/3=2.67m$

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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ 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?

1 Answer

Explanation:

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If an object is moving at $4 m s^{- 1}$ $4 ms^-1$ 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?