A small block of mass m slides down a ramp without friction from point `A\', and

Question

A small block of mass m slides down a ramp without friction from point `A', and then begins to climb a circular loop of radius R. When it reaches point `B' at the top of the loop, it has slowed down just enough so that it just barely loses contact with the track. Determine the kinetic energy of the block when it is at the top of the loop. Similarly, symbolically determine the height H required for this to occur. Enter your symbolic answer as a function of m, g, and R, plus physical constants such as 1.0, 0.5, or ?. As with all symbolic answers, do not ever include units or unit conversion factors.

Explanation / Answer

Initially the block is at height H therefore the block will have mgH
when it reaches the botton the potential energy get converte into the kinetic energy
so we can say
mgH = (1/2)mV2
H = V2/2g
Now when it reach the top of the loop then we can say that again this energy get converted into potential and kinetic
(1/2)mV2 = mg(2R) + (1/2)mv2
Since the block get detached from the surface so we can say that the kinetic energy is zero
therefore
(1/2)mV2 = mg(2R)
Hence we can say
mgH = mg2R
H = 2R

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