# If a #9# #kg# object moving at #9# #ms^-1# slows down to a halt after moving #81# #m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

The frictional coefficient

First step is to find the acceleration:

(the negative sign just indicates that the acceleration is in the opposite direction to the initial velocity)

Second step is to find the force causing the acceleration:

This is the magnitude of the frictional force. The third step is to find the normal force, which is just the weight force on the object:

The final step is to find the friction coefficient:

Frictional coefficients do not have units.

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To find the coefficient of kinetic friction, we can use the equation:

[ F_{\text{friction}} = \mu_k \times F_{\text{normal}} ]

Where:

- ( F_{\text{friction}} ) is the force of friction,
- ( \mu_k ) is the coefficient of kinetic friction,
- ( F_{\text{normal}} ) is the normal force.

The normal force (( F_{\text{normal}} )) is equal to the weight of the object, which is ( mg ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).

The force of friction can be calculated using Newton's second law:

[ F_{\text{friction}} = ma ]

Where:

- ( m ) is the mass of the object,
- ( a ) is the acceleration of the object.

Given:

- Mass ( m = 9 , \text{kg} )
- Initial velocity ( v_i = 9 , \text{m/s} )
- Final velocity ( v_f = 0 , \text{m/s} )
- Displacement ( d = 81 , \text{m} )

Using the equation of motion:

[ v_f^2 = v_i^2 + 2ad ]

Solving for acceleration ( a ):

[ a = \frac{{v_f^2 - v_i^2}}{{2d}} ]

[ a = \frac{{0 - (9^2)}}{{2 \times 81}} ]

[ a = -\frac{{81}}{{81}} ]

[ a = -1 , \text{m/s}^2 ]

Using Newton's second law:

[ F_{\text{friction}} = ma ]

[ F_{\text{friction}} = (9 , \text{kg}) \times (-1 , \text{m/s}^2) ]

[ F_{\text{friction}} = -9 , \text{N} ]

Since the object is slowing down, the force of friction opposes its motion.

The normal force is equal to the weight of the object:

[ F_{\text{normal}} = mg ]

[ F_{\text{normal}} = (9 , \text{kg}) \times (9.8 , \text{m/s}^2) ]

[ F_{\text{normal}} = 88.2 , \text{N} ]

Now, we can find the coefficient of kinetic friction:

[ \mu_k = \frac{{F_{\text{friction}}}}{{F_{\text{normal}}}} ]

[ \mu_k = \frac{{-9 , \text{N}}}{{88.2 , \text{N}}} ]

[ \mu_k ≈ -0.102 ]

The negative sign indicates that the force of friction acts in the opposite direction of motion. Therefore, the coefficient of kinetic friction is approximately ( 0.102 ).

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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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