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A block of inertia m is placed on an inclined plane that makes an angle θ with the horizontal. The block is given a shove directly up the plane so that it has initial speed v and the coefficient of kinetic friction between the block and the plane surface is μ. Part A

How far up the plane does the block travel before it stops?

Express your answer in terms of some or all of the variables m, \Theta, v, and \mu.

Part B

If the coefficient of static friction between block and surface is \mu, what maximum value of \Theta allows the block to come to a halt somewhere on the plane and not slide back down?

Express your answer in terms of some or all of the variables m, \Theta, v, and \mu.

Respuesta :

Answer:

A) d = v² / (2g (μ cos θ + syn θ)     B)    μ = tan θ

Explanation:

Part A

We can work this part with the work and energy theorem, where the work of the friction forces is equal to the energy change of the system.

The work is

       W = fr .d

With the force of friction it opposes the movement

       W = - fr d

The energy at the lowest point is

      Em₀ = K = ½ m v²

The energy at the highest point

      [tex]Em_{f}[/tex] = U = m g y

The height (y) can be found by trigonometry

      sin θ = y / d

      y = d sin θ

     W =  [tex]Em_{f}[/tex] –Em₀

     -fr d = mg d sin θ - ½ m v²

The equation for the force of friction is

      fr = μ N

From Newton's second law

      N - W cos Te = 0

We replace

     -μ (mg cos θ) d - mg d sin θ = - ½ m v²

      d g (μ cos θ + sin θ) = ½ v²

      d = v² / (2g (μ cos θ + syn θ)

Part B

The block is stopped, what is the Angle tet, let's use Newton's second law

      fr - W sin θ = 0       ⇒     fr = W sin θ

      N - W cos θ= 0       ⇒    N = w cos θ

      fr = μ N

      μ (mg cos θ) = mg syn θ

      μ = syn θ / cos θ

       μ = tan θ

jayilych4real

The distance the plane travels before it stops is d = v² / (2g (μ cos θ + syn θ)    

The maximum value of \Theta allows the block to come to a halt somewhere on the plane and not slide back down is μ = tan θ

Calculations and Parameters:

Using the work and energy theorem, we recall that the friction forces is equal to the change in energy of the system.

Hence, the work is W = fr .d

Recall that the force of friction opposes the movement

W = - fr d

The energy at the lowest point is

Em₀ = K = ½ m v²

The energy at the highest point

= U = m g y

The height (y) can be found by trigonometry

  • sin θ = y / d
  • y = d sin θ
  • W =   –Em₀
  • -fr d = mg d sin θ - ½ m v²

The equation for the force of friction is

fr = μ N

From the 2nd law of Newton

N - W cos Te = 0

Putting the values together,

  • -μ (mg cos θ) d - mg d sin θ = - ½ m v²
  • d g (μ cos θ + sin θ) = ½ v²
  • d = v² / (2g (μ cos θ + syn θ)

Solving for the Second Part

Using the 2nd law of Newton because the movement has stopped:

  • fr - W sin θ = 0       ⇒     fr = W sin θ
  • N - W cos θ= 0       ⇒    N = w cos θ
  • fr = μ N
  • =μ (mg cos θ) = mg syn θ
  • =μ = syn θ / cos θ

μ = tan θ

Read more about Newton's 2nd law here:

https://brainly.com/question/25545050

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