Two masses, m1 = 2.26 kg and m2 = 6.03 kg are on a horizontal frictionless surface and they are connected together with a rope as shown in the figure.
1. The rope will snap if the tension in it exceeds 75.0 N. What is the maximum value of the force F which can be applied?Given: - m1 = 2.26 kg- m2 = 6.03 kg- T = 75.0 NNeed: - max force in (N)Equations:- to find max forcemax force = tension + (acceleration*m2)- BUT we do not know acceleration so to find that, we use the equation:acceleration = tension/m1Work:acceleration = tension/m1a = 75.0N/2.26kga = 33.1858 m/s^2Now plug the acceleration into the max force equation:max force = T + a(m2)F(max) = 75.0N + (33.1858*6.03kg)F(max) = 275.2 N
2. What is the acceleration of the whole system, when this maximum force is applied?Equation: acceleration = F(max)/(m1+m2)a = 275.2N/(2.26kg+6.03kg)a = 33.2 m/s^2
Möchten Sie kostenlos Ihre eigenen Notizen mit GoConqr erstellen? Mehr erfahren.