If an object is moving at $50 \frac{m}{s}$ $50 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2 /g$ , how far will the object continue to move?

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If an object is moving at $50 \frac{m}{s}$ $50 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2 /g$ , how far will the object continue to move?

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Answer 1 · 2018-08-11T09:07:55

$625 m$ $625m$

Explanation:

Because of kinetic frictional force,the object will undergo constant deceleration.

So,to find upto what length it will go,we can apply the relation between velocity ( $v$ $v$ ),deceleration ( $a$ $a$ ) and displacement ( $s$ $s$ )

i.e $v^{2} = u^{2} - 2 a s$ $v^2=u^2-2as$ (where, $u$ $u$ is the initial velocity)

Now, frictional force acting is $f = μ m g = \frac{2}{g} \times m g = 2 m$ $f=mumg=2/g×mg=2m$ where, $m$ $m$ is its mass.

So, deceleration i.e $a = \frac{f}{m} = 2 m s^{- 2}$ $a=f/m=2ms^-2$

Putting in the equation and also putting $v = 0$ $v=0$ as it will travel until its final velocity becomes zero.

So,we get, $0^{2} = 50^{2} - 2 \times 2 \times s$ $0^2=50^2-2×2×s$

Or, $s = 625 m$ $s=625m$

Answer 2 · 2018-08-11T09:16:09

The distance is $= 625 m$ $=625m$

Explanation:

The coefficient of kinetic friction is

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

The mass of the object is $= m$ $=m$

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

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

Therefore,

$F_{r} = μ_{k} N = μ_{k} m g$ $F_r=mu_kN=mu_kmg$

According to Newton's Second Law

$F = m a$ $F=ma$

The acceleration is

$a = \frac{F}{m} = - \frac{F_{r}}{m} = - \frac{μ_{k} m g}{m} = - μ_{k} g$ $a=F/m=-F_r/m=-(mu_kmg)/m=-mu_kg$

The coefficient of kinetic friction is $μ_{k} = \frac{2}{g}$ $mu_k=2/g$

$a = - \frac{2}{g} \cdot g = - 2 m s^{- 2}$ $a=-2/g*g=-2ms^-2$

The initial velocity is $u = 50 m s^{- 1}$ $u=50ms^-1$

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

Apply the equation of motion

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

The distance is

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

$= \frac{0 - 50^{2}}{2 \cdot - 2}$ $=(0-50^2)/(2*-2)$

$= \frac{50^{2}}{4}$ $=50^2/4$

$= 625 m$ $=625m$

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If an object is moving at $50 \frac{m}{s}$ $50 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2 /g$ , how far will the object continue to move?

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Explanation:

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If an object is moving at $50 \frac{m}{s}$ $50 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{2}{g}$ $u_k=2 /g$ , how far will the object continue to move?