# If an object with a mass of #10 kg # is moving on a surface at #15 ms^-1# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?

The acceleration of the object is

First we calculate the acceleration of the object:

Rearranging:

The negative sign just shows that it is a deceleration, and we can ignore it for the rest of this calculation.

Now we find the force acting to slow the object, using Newton's Second Law :

Rearranging and substituting in the force:

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The friction coefficient of the surface is approximately 0.375.

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To find the friction coefficient of the surface, we can use the equation relating the force of friction to the mass of the object, its initial velocity, the time it takes to come to a stop, and the acceleration due to friction.

The equation for the force of friction is:

[ F_{\text{friction}} = m \cdot a ]

where:

- ( F_{\text{friction}} ) is the force of friction,
- ( m ) is the mass of the object,
- ( a ) is the acceleration due to friction.

We can find the acceleration due to friction using the equation of motion:

[ v = u + at ]

where:

- ( v ) is the final velocity (which is 0 m/s since the object comes to a halt),
- ( u ) is the initial velocity (given as 15 m/s),
- ( a ) is the acceleration due to friction,
- ( t ) is the time taken for the object to come to a halt (given as 4 s).

Rearranging the equation to solve for acceleration (( a )), we get:

[ a = \frac{v - u}{t} ]

Substitute the given values:

[ a = \frac{0 - 15}{4} = -3.75 \text{ m/s}^2 ]

Now, we can use the equation for force of friction:

[ F_{\text{friction}} = m \cdot a ]

Substitute the mass of the object (10 kg) and the calculated acceleration (-3.75 m/s^2):

[ F_{\text{friction}} = 10 \times (-3.75) = -37.5 \text{ N} ]

The force of friction is equal to the friction coefficient (( \mu )) multiplied by the normal force (( F_{\text{normal}} )). Since the object is coming to a halt, the force of friction is equal in magnitude to the applied force. Therefore, ( F_{\text{friction}} ) is equal to the coefficient of friction multiplied by the normal force.

Given that the normal force is equal to the weight of the object (( F_{\text{normal}} = m \cdot g )), where ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )), we can rearrange the equation to solve for the friction coefficient (( \mu )):

[ \mu = \frac{F_{\text{friction}}}{F_{\text{normal}}} = \frac{-37.5}{10 \times 9.8} ]

[ \mu = \frac{-37.5}{98} \approx -0.382 ]

The friction coefficient of the surface is approximately ( -0.382 ). Since friction coefficients are typically positive, this result may indicate an error in the calculation or a misunderstanding of the problem. Friction coefficients are usually represented as positive values.

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