Physics, 04.07.2019 08:00 karen7088

Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) find the angle of the slope down which this skier could coast at a constant velocity. you can neglect air resistance in both parts, and you will find the result of exercise 5.9 to be useful. explicitly show how you follow the steps in the problem-solving strategies.

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Acceleration of a skier heading down a 10.0º slope is 0.74 m/s² and the angle of the slope down which this skier could coast at a constant velocity is 5.7 degrees.

What is the acceleration of the object?

Acceleration of a object is the rate of change of velocity of the object per unit time.

Negative velocity of the object indicates the direction of object is backward, and positive velocity of the object indicates the direction of object is forward.

(a) The acceleration of a skier heading down a 10.0º slope-

From the second law of motion, the net force is equal to the mass times' acceleration. The net force on the body is due to own weight and friction force.

$W+F_f=ma\\mg\sin\theta+(-\mu mg\cos \theta)=ma\\a=g(\sin\theta-\mu \cos\theta)\\$ ........1

Put the values as,

$a=9.81[\sin(10)-0.1\times \cos (10)]\\a=0.74\rm m/s^2$

(b) The angle of the slope down which this skier could coast at a constant velocity-

For the constant velocity, the value of acceleration should be zero. Put the value of acceleration 0 in the equation 1 as,

$0=9.81(\sin\theta-0.1\times\cos\theta)\\\dfrac{\sin\theta}{\cos\theta}=0.1\\tan\theta=0.1\\\theta=5.7^o$

Acceleration of a skier heading down a 10.0º slope is 0.74 m/s² and the angle of the slope down which this skier could coast at a constant velocity is 5.7 degrees.

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Answer from: jessebeck0805

(a) 0.74 m/s^2

Explanation:

There are two forces acting on the skier: the component of the weight parallel to the slope, which acts downward, and the frictional force, which acts upward along the incline.

The component of the weight parallel to the inclined plane is:

$W= m g sin \theta$

where m is the mass of the skier, g=9.81 m/s^2 and $\theta=10^{\circ}$ .

The frictional force is instead

$F_f = -\mu m g cos \theta$

$\mu=0.1$ is the coefficient of friction for waxed wood on wet snow.

If we apply Newton's second law, we can write that the net force must be equal to the product of mass per acceleration:

$mgsin \theta -\mu mg cos \theta =ma$

And symplifying m, we can find the acceleration:

$a=g sin \theta-\mu g cos \theta=$

$=(9.81 m/s^2)(sin 10^{\circ})-(0.1)(9.81 m/s^2)(cos 10^{\circ})=0.74 m/s^2$

(b) $5.7^{\circ}$

Explanation:

This time, the skier is moving at constant velocity. Therefore, the acceleration is zero (a=0) and Newton's second law becomes:

$mg sin \theta - \mu m g cos \theta=0$

By simplifying, we get

$tan \theta = \mu$

From which we can find the angle at which the skier could coast at a constant velocity:

$\theta= tan^{-1} (\mu) = tan^{-1} (0.1)=5.7^{\circ}$

Answer from: Quest

Distance^2= yellow line ^2 (see picture) = 30^2 + 49^2so distance = squareroot of (300^2 +49^2)i dont know if this is how to solve it, but i cnt think of another way

Answer from: Quest

Aliquid is an answer i hope this

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