Friction Mechanics: Solving for Block's Speed at Point A

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SUMMARY

The discussion focuses on calculating the speed of a wooden block with a mass of 3.5 kg as it passes point A on its descent after sliding up a rough slope inclined at 29 degrees. The block initially moves at 20 m/s and comes to rest after traveling 25 m up the slope. The coefficient of friction was determined to be 0.554, leading to an acceleration of -8 m/s². The final calculations reveal that the speed of the block as it passes point A on the way down is derived from the equation a = -10[sin(29) + μcos(29)], resulting in a coefficient of friction μ = 0.36.

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Shah 72
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A wooden block of mass 3.5kg is sliding up a rough slope and passes a point A with speed 20m/s. The slope is at 29 degree to the horizontal. The block comes to rest 25m up the slope. Find its speed as it passes point A on the way down.
I calculated coefficient of friction = 0.554.
U=20m/s v= 0 and s= 25m.
By using v^2= u^2+2as, I got a= -8m/s^2.
I can't calculate after this.
Pls help
 
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upslope …

$a = \dfrac{0^2-20^2}{50} = -8 \, m/s^2 = -10[\sin(29)+\mu\cos(29)] \implies \mu = 0.36$

try again …
 

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