a block of mass is released from rest from a height. the block slides on a frictionless track and reaches the bottom. what is the velocity of the block when it reaches the bottom
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velocity of block when it reaches at bottom is. =√[2g*heigh]
where g is gravitational acceleration
where g is gravitational acceleration
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2
Since P.E =K.E
Therfore
mgh=1/2mv^2
gh.=1/2v^2
v^2=2gh
v=root of 2gh
Therfore
mgh=1/2mv^2
gh.=1/2v^2
v^2=2gh
v=root of 2gh
9292SAAD:
pls mark as brainlist ans.
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