Two bodies of masses 1, kg and 4, kg are conn | vedclass

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(C) Mass of the bigger body $M = 4\, kg$.Mass of the smaller body $m = 1\, kg$.The smaller mass $(m = 1\, kg)$ executes simple harmonic motion ($S$.$H

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Mass of the bigger body $M = 4\, kg$.Mass of the smaller body $m = 1\, kg$.The smaller mass $(m = 1\, kg)$ executes simple harmonic motion ($S$.$H$.$M$.) with angular frequency $\omega = 25\, rad/s$ and amplitude $A = 1.6\, cm = 1.6 	imes 10^{-2}\, m$.Since $\omega = \sqrt{\frac{K}{m}}$,we have $K = m\omega^2 = 1 	imes (25)^2 = 625\, N/m$.The force exerted by the system on the floor is the sum of the weight of the bigger body,the weight of the smaller body,and the spring force.The normal force $N$ exerted by the floor on the system is $N = Mg + mg + F_{spring}$.The spring force $F_{spring}$ varies as the smaller mass oscillates. The maximum force exerted on the floor occurs when the spring is at its maximum extension.The maximum spring force is $F_{max} = KA = 625 	imes 1.6 	imes 10^{-2} = 10\, N$.Thus,the maximum force exerted on the floor is $F_{total} = Mg + mg + F_{max} = (4 	imes 10) + (1 	imes 10) + 10 = 40 + 10 + 10 = 60\, N$.

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