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Arisa [49]
3 years ago
12

A hockey puck struck by a hockey stick is given an initial speed v0 in the positive x-direction. The coefficient of kinetic fric

tion between the ice and the puck is µk.
(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: µk and g.)
a =
(b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables v0, µk, and g only.
Physics
1 answer:
artcher [175]3 years ago
8 0

(a) a = -\mu_k g

In order to find the acceleration of the puck, we must analyze the forces acting on it.

Along the vertical direction, we have two forces: the weight of the puck, mg, and the normal reaction, N. The two forces are equal and opposite so that the vertical acceleration is zero, so we can write

N-mg = 0\rightarrow N = mg (1)

where m is the mass of the puck and g is the acceleration of gravity.

Along the horizontal direction, there is only one force acting on the puck: the force of friction, opposite to the direction of motion of the puck. So we can write:

-\mu_k N = ma (2)

where \mu_k is the coefficient of kinetic friction and a is the acceleration. Substituting (N) from eq.(1) into (2), we find an expression for a:

-\mu_k mg = ma\\a = -\mu_k g

(b) d=\frac{v_0^2}{2\mu_kg}

We can find the distance that the puck slides using the following SUVAT equation:

v^2-v_0^2 = 2ad

where

v = 0 is the final velocity of the puck (it comes to a stop, so v = 0)

v_0 is the initial velocity of the puck

a is the acceleration

d is the distance travelled

Substituting the expression for (a) that we found previously, we find an expression for d:

d=\frac{-v_0^2}{2a}=\frac{-v_0^2}{-2\mu_k g}=\frac{v_0^2}{2\mu_kg}

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Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc
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Answer:

a) 24.33 m of distance.

b) 34.55° east of the south.

Explanation:

The question is incomplete. The whole exercise is the following:

<em>"Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different directions. Ricardo walks 28.0 m in a direction 60.0° west of north. Jane walks 12.0 m in a direction 30.0° south of west. They then stop and turn to face each other. </em>

<em>(a) What is the distance between them? </em>

<em>(b) In what direction should Ricardo walk to go directly toward Jane?</em><em>"</em>

Now that we know what we need to do in this question, let's head for every part of the problem.

<u>a) Distance between Ricardo and Jane</u>

In this case, we need to analyze the given data:

Ricardo (which we will call R) is 28 m from the starting point at 60° west of north, and Jane (J) is 12 m at 30° south of west. So the distance between them, will be the point where they both stop and face each other. This point can be seen in the image attached (See picture).

Let's call the distance between them as "D", to get the distance of D, according to the picture will be:

D = J - R   (1)

However, as they are facing in different angles and directions, we cannot do the difference of their values distance just like that. In order to do that, we need to calculate the components in the "x" and "y" axis of each vector. In that way, we can get the components of x and y of the Distance D, and then, the whole distance between them will be:

D = √Dx² + Dy²     (2)

So, let's get the components of x and y of R and J.

For Ricardo (R):

Rx = R sin60° = 28 sin60° = -24.25 m

Ry = R cos60° = 28 cos60° = 14 m

The sign "-" it's because R it's on the second quadrant, therefore in x, we'll have to add the negative.

For Jane (J):

Jx = J cos30° = 12 cos30° = -10.39 m

Jy = J sin30° = 12 sin30° = -6 m

Again, the negative is added because J is on the third quadrant.

Now that we have the components, let's calculate vector D using expression (1):

Dx = -10.39 - (-24.25) = 13.86 m

Dy = -6 - 14 = -20 m

Now, using expression (2) we can finally know the distance between Jane And Ricardo:

D = √(-20)² + (13.86)²

<h2>D = 24.33 m</h2>

This is the distance between Jane and Ricardo.

b) Direction of Ricardo walking to Jane

In this case, we already have the components of x and y of the distance between them, so, to know the direction:

Tanα = Dy/Dx

α = tan⁻¹ (Dy/Dx)

Replacing the values we have:

α = tan⁻¹ (-20/13.86)

α = 55.45°

Which should south of east or:

β = 90 - 55.45

<h2>β = 34.55°</h2>

Ricardo should walk 34.55° east of south

Hope this helps

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