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))`.

Options

A. `2kg, mu_(s) = 0.4`
B. `4.2 kh, mu_(s) = 0.358`
C. `2.56 kg, mu_(s) ge 0.358`
D. `2.56 kg, mu_(s) le 0.358`

Answer

C

Explanation

`T = 2pi sqrt((m_(1)+m_(2))/(k))` and `mu =(A omega^(2))/(g)`