the problem
Consider a block of mass m sliding down the inclined plane shown below with a frictional force that is given by
- [math]F_f = kmv^2[/math]
200 px
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
200 px
Step 4: Define the Force vectors using the above coordinate system
- [math]\vec{N} = \left | \vec{N} \right | \hat{j}[/math]
- [math]\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]\vec{F_f} = - kmv^2 \hat{i}[/math]
Step 5: Used Newton's second law
Motion in the [math]\hat i[/math] direction described by Newton's second law is:
- [math]\sum F_{ext} = mg \sin \theta -mkv^2 = ma_x = m \frac{dv_x}{dt}[/math]
- Notice a terminal velocity [math]v_t[/math] exists when [math]a_x =0[/math]
- [math] mg \sin \theta -mkv^2 = ma_x = 0[/math]
- [math] \Rightarrow v_t^2 = \frac{g \sin \theta}{k}[/math]
Insert the terminal velociy constant into Newton's second law
- [math]\sum F_{ext} = mk \left ( v_t^2 - v^2 \right) = ma_x = m \frac{dv_x}{dt}[/math]
- [math]\int_0^t dt = \int_{v_i}^v \frac{dv}{k \left ( v_t^2 - v^2 \right)}[/math]
Integral table [math]\Rightarrow[/math]
- [math]\int \frac{dx}{a^2 + b^2x^2} = \frac{1}{ab} \tan^{-1} \frac{bx}{a}[/math]
- [math]a^2 = v_t^2 = \frac{g \sin \theta}{k}[/math]
- [math]b^2= -1[/math]
- [math]\int \frac{dv}{g\sin \theta - kv^2} = \frac{1}{\sqrt{-gk\sin \theta}} \tan^{-1} \left ( \sqrt{\frac{-k}{g \sin \theta}} \; v \right )[/math]
- [math]i \equiv \sqrt{-1}[/math]
- [math]\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]i\tan^{-1}(icx) = -\tanh^{-1}(cx) = -\tanh^{-1}\left (\frac{\left | b \right |}{a} x\right )[/math]
Identities
- [math]\tan^{-1}(z) = \frac{i}{2} \log \left ( \frac{i + z}{i-z}\right )[/math]
- [math]\tanh^{-1}(z) = \frac{1}{2} \log \left ( \frac{1 + z}{1-z}\right )[/math]
- [math]\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]t = \frac{1}{\sqrt{gk\sin \theta}} \tanh^{-1} \left ( \sqrt{\frac{k}{g \sin \theta}} \; \;v \right )[/math]
Solving for [math]v[/math]
- v = \tan \left ( \sqrt{gk\sin \theta} i t \right )
- =
Forest_UCM_NLM#Block_on_incline_with_friction
