Physics, 07.07.2019 21:00 KoolCatz

Askier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along this slope before coming to rest. part a what was the average coefficient of friction?

Answers: 1

Show answers

Answers

The attached free-body diagram shows the forces that are responsible for the skier coming to rest eventually on the incline.

We see that the component of his Weight along the incline $mgSin \alpha$ and the Friction both act in tandem to stop him.

In order to calculate the Friction, we can make use of Newton's 2nd law, which states that $F_{net} = ma$

The $F_{net}$ here is given by $mgSin \alpha + F_{k}$ , where $F_{k}$ is the Kinetic Friction.

We also know that the magnitude of Friction Force can be calculated using the equation $F_{k} =$ μ. $F_{N}$ , where $F_{N}$ is the Normal Force acting perpendicular to the incline as shown in the figure.

We see that $F_{N} = mgCos \alpha$ since they both balance each other out.

Hence, putting all these together, we have $mgSin \alpha +$ μ. $mgCos \alpha = ma$

Simplifying this, we get $gSin \alpha +$ μ $gCos \alpha = a$

We clearly see that we need to calculate the acceleration before we can obtain the value of the Coefficient of Friction μ

And for that, we make use of the following data obtained from the question:

Initial Velocity $V_{i} = 11.0 m/s$

Final Velocity $V_{f} = 0$

Displacement along the incline D = 15m

Acceleration a = ?

Using the equation $V_{f} ^{2} = V_{i} ^{2} + 2aD$ , and

Plugging in known numerical values, we get $0 = (11)^{2} + 2a(15)$

Solving for a gives us, $a = -4.03 m/s^{2}$

Since the negative sign indicates that this is deceleration, we can ignore the sign and consider the magnitude alone.

Thus, plugging in $a = 4.03 m/s^{2}$ in the force equation we wrote above, we have

(9.8)Sin (17) + μ. (9.8)Cos (17) = 4.03

Solving this for μ, we get its value as μ = 0.124

Thus, the average coefficient of friction on the incline is 0.12

Askier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along

Answer from: Quest

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

-reverse card

Answer from: Quest

Year 1936, lehman predicted that the earths 'inner core' is solid. she observed that the seismic signals during her study were bent and did not pass through instead bouncing back. later it was confirmed that the earth's inner core was solid made up of nickel-iron and had a molten outer core.

Another question on Physics

Physics, 21.06.2019 13:30

What do butterflies have on their feet which allows them to sense different strengths and types of nectar? a.mechanoreceptors b.acoustic receptors c.substratum receptors d.chemoreceptors

Answers: 2

Answer

Physics, 22.06.2019 16:00

The field between two charged parallel plates is kept constant. if the two plates are brought closer together, the potential difference between the two plates either a) decrease b) does not change c) increase?

Answers: 3

Answer

Physics, 22.06.2019 21:30

The diagram shows a boulder rolling down a hill into a valley and then up the opposite hill. at which position does the boulder have the greatest kinetic energy? ab c d

Answers: 2

Answer

Physics, 23.06.2019 00:30

Which change would make the diagram correct?

Answers: 1

Answer

You know the right answer?

Askier traveling 11.0 m/s reaches the foot of a steady upward 17 â incline and glides 15 m up along...

Questions