# If a #1/24 kg# object moving at #5/4 m/s# slows to a halt after moving #1/6 m#, what is the coefficient of kinetic friction of the surface that the object was moving over?

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To find the coefficient of kinetic friction ((\mu_k)), you can use the equation:

[\mu_k = \frac{F_{friction}}{N}]

Where: (F_{friction}) = force of friction (N) = normal force

First, find the force of friction using the equation:

[F_{friction} = m \times a]

Where: (m) = mass of the object (a) = acceleration

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

Where: (v_f) = final velocity (0 m/s, as the object comes to a halt) (v_i) = initial velocity (d) = displacement

Then, calculate the normal force ((N)) using the equation:

[N = m \times g]

Where: (g) = acceleration due to gravity (approximately 9.8 m/s^2)

Finally, substitute (F_{friction}) and (N) into the equation for (\mu_k):

[\mu_k = \frac{m \times a}{m \times g}]

Given: (m = \frac{1}{24}) kg (v_i = \frac{5}{4}) m/s (d = \frac{1}{6}) m

[a = \frac{{0 - (\frac{5}{4})^2}}{{2 \times (\frac{1}{6})}}] [N = \frac{1}{24} \times 9.8]

Then solve for (\mu_k).

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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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