A block of mass m is placed on the top of a 6 kg cart such that the time period of the system is 0.75 s assuming there is no slipping. If the cart is displaced by 50 mm from its equilibrium position and released, then the coefficient of static friction μ s between block and cart is just sufficient to prevent the block from sliding. The value of m and μ s respectively are (Take g = 9 .8 m / s 2 )

Step-by-step solution

T = 2 π m + 6 600 ( T = 2 π m k ) or 0.75 = 2 π m + 6 600 ∴ m = 0.75 2 × 600 2 π 2 - 6 = 2.55 kg Maximum acceleration of SHM is, a max = ω 2 A (A = amplitude) i.e., maximum force on mass 'm' is m ω 2 A which is being provided by the force of friction between the mass and the cart. Therefore, μ s mg ≥ mω 2 A or μ s ≥ ω 2 A g or μ s ≥ ( 2 π T ) 2 · A g or μ s ≥ ( 2 π 0.75 ) 2 ( 0.05 9.8 ) A = 50 mm or μ s ≥ 0.358 Thus, the minimum value of μ s should be 0.358.