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Almoore01

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Hi, here's another problem I'm a little stuck on:

For part A I just used Conservation of Energy and got the maximum height "H" to be H = (((Vi)^2)/(2g)), which is the right answer.

For part B, I know I have to use Work = (Change in K + Change in U). So I have:

(umg(cos (theta))) * distance = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (mgHf - mgHi). The masses cancelled out and I assumed that the initial height and initial and final velocities equaled 0, so then my answer ended up being Hf = ucos(theta) *d, which isn't the right answer...

Any suggestions?

**A block of mass M slides up an incline with an initial speed Vi in the position shown. (The figure is of a block sliding up an incline at an angle "theta" with a final height of "H."**

Part A: If the incline in fritionless, determine the maximum height H to which the block will rise, in terms of the given quantities and appropriate constants.

Part B: If the incline is rough with a coefficient of sliding friction "u" (mew), determine the maximum height to which the block will rise in terms of "H" and given quantities.Part A: If the incline in fritionless, determine the maximum height H to which the block will rise, in terms of the given quantities and appropriate constants.

Part B: If the incline is rough with a coefficient of sliding friction "u" (mew), determine the maximum height to which the block will rise in terms of "H" and given quantities.

For part A I just used Conservation of Energy and got the maximum height "H" to be H = (((Vi)^2)/(2g)), which is the right answer.

For part B, I know I have to use Work = (Change in K + Change in U). So I have:

(umg(cos (theta))) * distance = ((1/2)((Vf)^2) - (1/2)((Vi)^2)) + (mgHf - mgHi). The masses cancelled out and I assumed that the initial height and initial and final velocities equaled 0, so then my answer ended up being Hf = ucos(theta) *d, which isn't the right answer...

Any suggestions?

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