Wood block coeffiecent of friction

AI Thread Summary
The discussion revolves around calculating the distance a wooden block slides up an incline and its speed when returning to the starting point, given an initial speed of 6 m/s and a coefficient of friction of 0.3 on a 25-degree incline. The energy method suggests that without friction, the height reached would be 1.83 m, but accounting for friction leads to a calculated height of approximately 1.065 m. The acceleration down the slope is determined to be 7.15 m/s², leading to a time of 0.84 seconds to return to the starting point. The final speed on the way down must consider friction, which significantly affects the overall calculations. Accurate accounting for forces and friction is crucial for solving the problem effectively.
Frosty_TAW
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I also need help with this problem, it involves energy and i just can't crack the problem by using what i know, i can't even start it, so here it is:
A wooden Block whose intitial speed is 6 m/s starts to slide up an incline plane at 25 degrees above the horizontal.
a)if the coeffiecent of friction is .3 how far up the plane does the block go?
b) what will be the blocks speed after it has slid back down the plane to its starting point?
 
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a)the energy way is this but i think that assumes no friction
\frac{v^2}{2g} = h

u have to add up all the forces on the block |ma| = |F_g + F_f + N|
which if u know vectors i think gives...
a = 7.15 \frac{m}{s^2} down the slope
so the block makes it v = v_0 + at, 0 = 6 - 7.15 t, t = 0.84s and therefore
x = 6(.84s) - 1/2 (7.15) (.84s)^2 = 2.52 m which is the hypotenus
h = 2.52m sin 25 = 1.065 m so you i put a lot of round off error so check it out
(the energy equation yeilds h = 1.83 m so without the friction its off 58%!)

b) just make sure u account for the friction on the way down
 
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thanks man, that helps alot
 
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