A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho uld a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?
1 answer:
Answer:
same 0.81m
Explanation:
in this problem if we assume there no resistance of any sort. and we apply the energy conservation
change in Potential energy = change in kinetic energy
mgh = 0.5mv^2
gh = 0.5v^2
the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.
so at the bottom
put h = 0.81m
9.81 * 0.81 * 2 = v^2
v=3.99 m/s
both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size
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