With the assumption of no slipping, determine the mass m of the block which must be placed on the top of a `6 kg` cart in order that the system period is `0.75s`. What is the minimum coefficient of static friction `mu_(s)` for which the block will not slip relative to the cart is displaced `50 mm` from the equilibrium position and released? Take `(g = 9.8 m//s^(2))`.

Answer

A, B, C

Explanation

(a) `T = 2pisqrt((m+6)/(600))" "(T = 2pi sqrt((m)/(K)))`
or `0.75 = 2pisqrt((m+6)/(600))`
`:. m = ((0.75)^(2)xx600)/((2pi)^(2))-6`
`= 2.55kg `
(b) Maximum acceleration of SHM is,
`a_(max) = omega^(2)A` (A = amplitude)
i.e. maximum force on mass 'm' is `m omega^(2)A` which is bieng provided by the force of friction between the mass and the cart. Therefore,
`mu_(s)mg ge momega^(2)A or mu_(s) ge (omega^(2)A)/(g)` or ` mu_(s) ge ((2pi)/(T))^(2). A/g`
or `mu_(s) ge((2pi)/(0.75))^(2) ((0.05)/(9.8))`
`(A = 50 mm)`
or `mu_(s)ge0.358`
Thus, the minimum value of mu_(s) should be `0.358`.