A solid sphere of mass 2 kg rolls on a horizontal surface at 10m/s.it then rolls up a rough inclined plane of inclination 30 with horizontal.the height attained by sphere before it stops is?
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Answered by
11
Mechanical energy is conserved.
u = initial velocity
v = final velocity = 0 (for max. height)
h = initial height
d = final height
m = mass
g = 10 m/s^2
(1/2)mu^2 + mgh = (1/2)mv^2 + mgd
dividing by m on both sides and substituting values,
(1/2)10^2 + 10x0 = (1/2)0^2 + 10d
=> 10d = 100/2
=> d = 10/2 = 5 meters
Answered by
5
Mechanical energy is conserved.
u = initial velocity
v = final velocity = 0 (for max. height)
h = initial height
d = final height
m = mass
g = 10 m/s^2
(1/2)mu^2 + mgh = (1/2)mv^2 + mgd
dividing by m on both sides and substituting values,
(1/2)10^2 + 10x0 = (1/2)0^2 + 10d
=> 10d = 100/2
=> d = 10/2 = 5 meters
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