Block launching off ramp given initial velocity and Friction

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SUMMARY

The discussion focuses on calculating the distance a block will land after being launched off a ramp inclined at 35 degrees, with a length of 5 meters and friction coefficients of 0.6 (static) and 0.2 (kinetic). The block is initially given a velocity of 12 m/s. The final velocity of the block as it exits the ramp is determined using the equation 0.5 * m * v^2 = m * h * g + m * g * cos(theta) * Kf * L + 0.5 * m * V^2. The correct distance from the ramp where the block lands is 9.75 meters, while the initial calculations yielded incorrect results around 8.57 meters due to misapplication of friction in the equations.

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physicN00Bz
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Homework Statement



a block is at the base of a ramp of angle 35degrees. The ramp is 5 m long and has a coefficient of static and kinetic friction of .6 and .2 respectively. The block is initially given a velocity of 12 m/s up the ramp, how far from the ramp will the block land?

my main problem is finding the final velocity as the block exit the ramp, honestly i really don't know where to start.

Homework Equations


to find the final velocity of the block exiting the ramp i use

0.5 * m * v^2 = m * h * g + m * g * cos (theta) * Kf * L + 0.5 * m * V^2


The Attempt at a Solution



The answer is 9.75 m, all of my answer come out to be 8.57m or something close, what am i doing wrong? how do i applied the static friction into the equation?
 
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physicN00Bz said:
to find the final velocity of the block exiting the ramp i use

0.5 * m * v^2 = m * h * g + m * g * cos (theta) * Kf * L + 0.5 * m * V^2

This equation gives the final speed the block leaves the ramp. How do you proceed to get the the distance from the ramp the block will land?

ehild
 

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