Posted by **tony** on Wednesday, October 20, 2010 at 2:31am.

A 1.9 kg block slides down a frictionless ramp, as shown in figure 8-28. The top of the ramp is 1.5 m above the ground; the bottom of the ramp is .25 m above the ground. The block leaves the ramp moving horizontally, and lands a horizontal distance d away. Find the distance d.

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**drwls**, Wednesday, October 20, 2010 at 6:46am
At the bottom of the ramp, the block has acquired kinetic energy MgH, where H = 1.5 - 0.25 = 1.25 m.

Its velocity is therefore

V = sqrt(2gH) = 4.95 m/s

Multiply that (horizontal) velocity by the time it takes to fall H' = 0.25 m vertically.

T = sqrt (2H'/g) = 0.051 s

That will give you the distance d.

d = sqrt (2gH) * sqrt (2H'/g)

= 2 sqrt(H*H')

It is interesting that d is independent of g. If g were zero, it would not fall down the ramp at all

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**Amir**, Thursday, November 11, 2010 at 11:22pm
A 2 kg box slides down a 25 degree ramp with an acceleration of 2 meters per second. Determine the coefficient of kinetic friction between the box and the ramp?

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