It was a snow day, so Eric and some friends went sledding. He boarded his 60-inch sled and descended the steepest section near the tall trees. Before his ride ended at the location of a tree stump buried a few inches below the snow, he was accelerating at 2.48 m/s/s down the 21.9° incline. Determine the coefficient of friction between the sled and the snow. (Coefficients of friction are positive, so use the absolute value of your answer.)Use the approximation g ≈ 10 m/s^2.Answer: ___________ (no units) (rounded to the thousandths place)

Question

03-11-2022
Physics

Answered

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It was a snow day, so Eric and some friends went sledding. He boarded his 60-inch sled and descended the steepest section near the tall trees. Before his ride ended at the location of a tree stump buried a few inches below the snow, he was accelerating at 2.48 m/s/s down the 21.9° incline. Determine the coefficient of friction between the sled and the snow. (Coefficients of friction are positive, so use the absolute value of your answer.)Use the approximation g ≈ 10 m/s^2.Answer: ___________ (no units) (rounded to the thousandths place)

Sagot :

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ShailynnB164156idn ShailynnB164156idn · Answer 1 · 2022-11-03T21:47:05-04:00

Answer:

0.335

Explanation:

We can make the following free body diagram

So, we can calculate the normal force as

Fn = mgcos(21.9)

Because the net force in the perpendicular axis is 0. Then, the friction force is equal to

Ff = μFn

Ff = μmgcos(21.9)

Where μ is the coefficient of friction.

Now, by the second law of newton, we have the following equation

Fnet = mgsin(21.9) - Ff = ma

Then

mgsin(21.9) - μmgcos(21.9) = ma

Now, we can solve the equation for μ, so

[tex]\begin{gathered} g\sin(21.9)-\mu g\cos(21.9)=a \\ \\ -\mu g\cos(21.9)=a-g\sin(21.9) \\ \\ \mu=\frac{a-g\sin(21.9)}{-g\cos(21.9)} \end{gathered}[/tex]

So, replacing a = 2.48 m/s² and g = 10 m/s², we get:

[tex]\mu=\frac{2.48-10\sin(21.9)}{-10\cos(21.9)}=0.335[/tex]

Therefore. the coefficient of friction is 0.335

Sagot :

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