Question

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.

Answer 1

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

      $Em_{f}$ = U = m g y

The height (y) can be found by trigonometry

      sin θ = y / d

      y = d sin θ

     W =  $Em_{f}$ –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