The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:
[tex]P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)[/tex]
Using the Velocity Hourly Equation, we get:
[tex]V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}[/tex]
Uniting the equations:
[tex]\Delta t= \frac{V}{g(sen\O-ucos\O)} [/tex]
Entering the unknowns:
[tex]\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)} \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)} \\ \Delta t= \frac{12}{5.27} \\ \boxed {\Delta t=2.28s}[/tex]
Obs: Approximate results
If you notice any mistake in my english, please let me know, because i am not native.
