# An object, previously at rest, slides #8 m# down a ramp, with an incline of #pi/3 #, and then slides horizontally on the floor for another #3 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Here length of the ramp (l)= 8m

Now applying conservation of mechanical energy we can write

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To find the coefficient of kinetic friction, we can use the work-energy principle. The work done by friction is equal to the change in kinetic energy of the object.

The change in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy. Since the object starts from rest, the initial kinetic energy is 0. The final kinetic energy is calculated using the formula ( KE = \frac{1}{2}mv^2 ), where ( m ) is the mass of the object and ( v ) is its final velocity.

The work done by friction is given by the formula ( W_{\text{friction}} = f \cdot d ), where ( f ) is the frictional force and ( d ) is the distance over which the force acts.

The frictional force can be calculated using ( f = \mu \cdot N ), where ( \mu ) is the coefficient of kinetic friction and ( N ) is the normal force. The normal force is equal to the weight of the object, which can be calculated as ( mg ), where ( g ) is the acceleration due to gravity.

The total distance traveled by the object is 8m down the ramp and 3m horizontally on the floor, for a total distance of 11m. The work done by friction is therefore ( f \cdot 11 ). Setting this equal to the change in kinetic energy, we have:

[ f \cdot 11 = \frac{1}{2}mv^2 ]

Substituting ( f = \mu \cdot mg ), we get:

[ \mu \cdot mg \cdot 11 = \frac{1}{2}mv^2 ]

The mass ( m ) cancels out, and we can solve for ( \mu ):

[ \mu \cdot g \cdot 11 = \frac{1}{2}v^2 ]

[ \mu = \frac{v^2}{2g \cdot 11} ]

To find ( v ), we can use the equation of motion:

[ v^2 = u^2 + 2as ]

where ( u ) is the initial velocity (0 in this case), ( a ) is the acceleration, and ( s ) is the total distance. The acceleration down the ramp can be calculated using ( a = g \cdot \sin(\pi/3) ), and the acceleration on the horizontal floor is 0 since there is no net force in that direction. Substituting these values, we get:

[ v^2 = 0 + 2 \cdot g \cdot \sin(\pi/3) \cdot 8 ]

[ v^2 = 2g \cdot \sin(\pi/3) \cdot 8 ]

[ v^2 = 2g \cdot \frac{\sqrt{3}}{2} \cdot 8 ]

[ v^2 = 8g\sqrt{3} ]

Finally, substituting this into the equation for ( \mu ), we get:

[ \mu = \frac{8g\sqrt{3}}{2g \cdot 11} ]

[ \mu = \frac{4\sqrt{3}}{11} ]

Therefore, the coefficient of kinetic friction is ( \frac{4\sqrt{3}}{11} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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