Two 5 kg masses are attached to opposite ends of a long massless cord which passes tautly over a massless frictionless pulley. The upper mass is initially held at rest on a table 50 cm from the pulley. The coefficient of kinetic friction between this mass and the table is 0.60. When the system is released, what's its resulting acceleration
+33654 Assuming there is no static friction (!) then the resistive force of the table on the upper mass is 0.6*5*9.8 Newtons.
The force on the lower mass is 5*9.8 Newtons.
The net force on the system is therefore 5*9.8 - 0.6*5*9.8 Newtons, or 0.4*5*9.8 Newtons.
The total mass of the system is 10kg, so, from Newton's second law of motion, the acceleration is net force/mass:
acceleration=0.4×5×9.810⇒acceleration=1.96
acceleration = 1.96 m/s2
.
+33654 Assuming there is no static friction (!) then the resistive force of the table on the upper mass is 0.6*5*9.8 Newtons.
The force on the lower mass is 5*9.8 Newtons.
The net force on the system is therefore 5*9.8 - 0.6*5*9.8 Newtons, or 0.4*5*9.8 Newtons.
The total mass of the system is 10kg, so, from Newton's second law of motion, the acceleration is net force/mass:
acceleration=0.4×5×9.810⇒acceleration=1.96
acceleration = 1.96 m/s2
.
