Forest UCM NLM BlockOnIncline

the problem

Consider a block of mass m sliding down the inclined plane shown below with a frictional force that is given by

[math]\displaystyle{ F_f = kmv^2 }[/math]

File:TF UCM InclinedPlaneWfriction.png

Find the blocks speed as a function of time.

Step 1: Identify the system

The block is the system with the following external forces, A normal force, a gravitational force, and the force of friction.

Step 2: Choose a suitable coordinate system

A coordinate system with one axis along the direction of motion may make solving the problem easier

Step 3: Draw the Free Body Diagram

File:TF UCM FBD InclinedPlaneWfriction.png

Step 4: Define the Force vectors using the above coordinate system

[math]\displaystyle{ \vec{N} = \left | \vec{N} \right | \hat{j} }[/math]

[math]\displaystyle{ \vec{F_g} = \left | \vec{F_g} \right | \left ( \sin \theta \hat{i} - \cos \theta \hat{j} \right )= mg \left ( \sin \theta \hat{i} - \cos \theta \hat{j} \right ) }[/math]

[math]\displaystyle{ \vec{F_f} = - kmv^2 \hat{i} }[/math]

Step 5: Used Newton's second law

Motion in the [math]\displaystyle{ \hat i }[/math] direction described by Newton's second law is:

[math]\displaystyle{ \sum F_{ext} = mg \sin \theta -mkv^2 = ma_x = m \frac{dv_x}{dt} }[/math]

Notice a terminal velocity [math]\displaystyle{ v_t }[/math] exists when [math]\displaystyle{ a_x =0 }[/math]

[math]\displaystyle{ mg \sin \theta -mkv^2 = ma_x = 0 }[/math]

[math]\displaystyle{ \Rightarrow v_t^2 = \frac{g \sin \theta}{k} }[/math]

Insert the terminal velociy constant into Newton's second law

[math]\displaystyle{ \sum F_{ext} = mk \left ( v_t^2 - v^2 \right) = ma_x = m \frac{dv_x}{dt} }[/math]

[math]\displaystyle{ \int_0^t dt = \int_{v_i}^v \frac{dv}{k \left ( v_t^2 - v^2 \right)} }[/math]

Integral table [math]\displaystyle{ \Rightarrow }[/math]

[math]\displaystyle{ \int \frac{dx}{a^2 + b^2x^2} = \frac{1}{ab} \tan^{-1} \frac{bx}{a} }[/math]

[math]\displaystyle{ a^2 = v_t^2 = \frac{g \sin \theta}{k} }[/math]

[math]\displaystyle{ b^2= -1 }[/math]

[math]\displaystyle{ \int \frac{dv}{g\sin \theta - kv^2} = \frac{1}{ikv_t} \tan^{-1} \left ( \frac{-v}{v_t}} \; v \right ) }[/math]

[math]\displaystyle{ i \equiv \sqrt{-1} }[/math]

[math]\displaystyle{ \int \frac{dv}{g\sin \theta - kv^2} = \frac{1}{\sqrt{gk\sin \theta}} i \tan^{-1} \left ( \sqrt{\frac{k}{g \sin \theta}} \; \;iv \right ) }[/math]

[math]\displaystyle{ i\tan^{-1}(icx) = -\tanh^{-1}(cx) = -\tanh^{-1}\left (\frac{\left | b \right |}{a} x\right ) }[/math]

Identities

[math]\displaystyle{ \tan^{-1}(z) = \frac{i}{2} \log \left ( \frac{i + z}{i-z}\right ) }[/math]

[math]\displaystyle{ \tanh^{-1}(z) = \frac{1}{2} \log \left ( \frac{1 + z}{1-z}\right ) }[/math]

[math]\displaystyle{ \tan^{-1}(ix) = \frac{i}{2} \log \left ( \frac{i + ix}{i-ix}\right )=\frac{i}{2} \log \left ( \frac{1 + 1x}{1-x}\right ) = i\tanh^{-1}(x) }[/math]

[math]\displaystyle{ t = \frac{1}{\sqrt{gk\sin \theta}} \tanh^{-1} \left ( \sqrt{\frac{k}{g \sin \theta}} \; \;v \right ) }[/math]

Solving for [math]\displaystyle{ v }[/math]

v = \tan \left ( \sqrt{gk\sin \theta} i t \right )

=

Forest_UCM_NLM#Block_on_incline_with_friction