Question

d) all result in the same final speed 3. Given: Mass of block:m R. Starting: rest at height When the object is at the top of the track shown by tick mark, it pushes against the track with a force equal to four times it's weight. Find: What height was the object dropped from?

Answer 1

from initial to final position,

change in height = h-2R

so, mg(h-2R) = 1/2 mv²

or, v = $\sqrt{2g(h-2R)}$

at the topmost poit of the track(arrow)

mg+N = mv²/R ....................centripetal force

N=4mg (given)

5mg = mv²/R

v = $\sqrt{5gR}$

so equating the two valus of v

5gR = 2g(h-2R)

5R=2h-4R

h=4.5R