A block slides down an inclined plane of slope angle ? with constant velocity. It is then projected up the same plane with an initial velocity 'u' . How far up the incline will it move before coming to rest?

Question

vikas askiitian expert · Answer

block moves down the plane with constant velocity it means net force is zero...

force due to friction (f) = mgsin@ (@ is the angle of plane)

net retardation force when body is projected upward is (f+mgsin@)

f = mgsin@ so

net retardation force = 2mgsin@

ma = mgsin@

a(retardation) = 2gsin@ .............1

V² = U² - 2aS

finally velocity becomes 0 so

U² = 2aS

S = U²/2a

=u²/4gsin@

Rishi Sharma · Answer

Dear Student,
Please find below the solution to your problem.

block moves down the plane with constant velocity it means net force is zero...
force due to friction (f) = mgsin@ (@ is the angle of plane)
net retardation force when body is projected upward is (f+mgsin@)
f = mgsin@ so
net retardation force = 2mgsin@
ma = mgsin@
a(retardation) = 2gsin@ .............1
V2 = U2 - 2aS
finally velocity becomes 0 so
U2 = 2aS
S = U2/2a
=u2/4gsin@

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A block slides down an inclined plane of slope angle ? with constant velocity. It is then projected up the same plane with an initial velocity 'u' . How far up the incline will it move before coming to rest?

A block slides down an inclined plane of slope angle ? with constant velocity. It is then projected up the same plane with an initial velocity 'u' . How far up the incline will it move before coming to rest?

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