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#### johnherald

##### New member

- Jan 10, 2019

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- Jan 10, 2019

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- Mar 1, 2012

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distance up the incline (with kinetic friction present) ...

$\Delta x = \dfrac{v_f^2 - v_0^2}{2a}$

time up the incline ...

$t = \dfrac{v_f-v_0}{a}$

... where $v_f=0$ and $a = -g(\sin{\theta} + \mu \cos{\theta})$

time down the incline ...

$\Delta x = v_0 \cdot t + \dfrac{1}{2}at^2 \implies t = \sqrt{\dfrac{2\Delta x}{a}}$

note $v_0 = 0$, $\Delta x$ is the opposite of that found going up the incline and $a = -g(\sin{\theta} - \mu \cos{\theta})$

- Jul 18, 2013

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- Jan 10, 2019

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and i have the difficulty solving some problems specially to the questions without a value.. thank you again for your help..