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

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To determine the speed of a wooden block sliding down a rough slope, the block's mass is 3.5 kg, and it initially moves up at 20 m/s before coming to rest 25 m up the slope, which is inclined at 29 degrees. The coefficient of friction was calculated to be 0.554, leading to an acceleration of -8 m/s². Using the equation v² = u² + 2as, the acceleration was further analyzed, resulting in a recalculated coefficient of friction of 0.36. The discussion emphasizes the need for accurate calculations to find the block's speed at point A on its return. The focus remains on solving the mechanics of motion under frictional forces.
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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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