If a #7/2 kg# object moving at #9/4 m/s# slows to a halt after moving #3/8 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?
While sliding, the vertical forces (gravitational and normal) cancel out, so we'll look at only horizontal motion to find the friction.
We know:
Now that we know the acceleration, we can use Newton's second law to find the magnitude of the friction force:
The coefficient of kinetic friction is thus
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To find the coefficient of kinetic friction, we can use the formula:
[ \text{Coefficient of kinetic friction} = \frac{\text{Force of friction}}{\text{Normal force}} ]
The force of friction can be calculated using Newton's second law:
[ \text{Force of friction} = \text{Mass} \times \text{Acceleration} ]
The acceleration can be calculated using the equation of motion:
[ v^2 = u^2 + 2as ]
Where: ( v ) = final velocity (0 m/s, since the object comes to a halt) ( u ) = initial velocity (( \frac{9}{4} ) m/s) ( a ) = acceleration ( s ) = displacement (( \frac{3}{8} ) m)
Solve for acceleration (( a )), then use it to find the force of friction. Finally, divide the force of friction by the normal force to get the coefficient of kinetic friction.
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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.
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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