posted by tony on .
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.
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
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?