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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?
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@
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@
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@ sonet retardation force = 2mgsin@ ma = mgsin@ a(retardation) = 2gsin@ .............1V2 = U2 - 2aSfinally velocity becomes 0 so U2 = 2aSS = U2/2a =u2/4gsin@Thanks and Regards
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