The masses mA =1.0 kg and m = 1.1 kg slide on the smooth (frictionless) triangular...

 

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The masses mA =1.0 kg and m = 1.1 kg slide on the smooth (frictionless) triangular block as shown in the figure. The pulley and the cord have a negligible mass. The triangular block is fixed to the bottom. sin A = 0.60, cosA=0.80,sinB = 0.50,COS OB 0,87 and g = 10 m/s, What is the acceleration of the object of mass me in units of m/s ? (a) 5.75 left upwards (b) 0.24 left upwards (c) 5 right downwards What is the tension on the cord approximately? (a) 5.76 N (b) 11 N (c) 6.2 N (d) 11.75 N (e) 12 N m e (d) 5 left upwards (e) 5.75 right upwards What is the vertical component (the direction of g) of the force acting on the triangular block due to ma ? (a) 10 N (b) 5 N (c) 6.4 N (e) 8 N (d) 9 N m 0 The masses mA =1.0 kg and m = 1.1 kg slide on the smooth (frictionless) triangular block as shown in the figure. The pulley and the cord have a negligible mass. The triangular block is fixed to the bottom. sin A = 0.60, cosA=0.80,sinB = 0.50,COS OB 0,87 and g = 10 m/s, What is the acceleration of the object of mass me in units of m/s ? (a) 5.75 left upwards (b) 0.24 left upwards (c) 5 right downwards What is the tension on the cord approximately? (a) 5.76 N (b) 11 N (c) 6.2 N (d) 11.75 N (e) 12 N m e (d) 5 left upwards (e) 5.75 right upwards What is the vertical component (the direction of g) of the force acting on the triangular block due to ma ? (a) 10 N (b) 5 N (c) 6.4 N (e) 8 N (d) 9 N m 0

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To solve this problem we need to apply Newtons second law of motion and consider the forces acting on the system such as tension and gravity Ill walk you through how to find the acceleration of mass m... View the full answer