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

Guest Jan 1, 2015

#1**+10 **

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:

$${\mathtt{acceleration}} = {\frac{{\mathtt{0.4}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{{\mathtt{10}}}} \Rightarrow {\mathtt{acceleration}} = {\mathtt{1.96}}$$

acceleration = 1.96 m/s^{2}

.

Alan
Jan 4, 2015

#1**+10 **

Best Answer

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:

$${\mathtt{acceleration}} = {\frac{{\mathtt{0.4}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{9.8}}}{{\mathtt{10}}}} \Rightarrow {\mathtt{acceleration}} = {\mathtt{1.96}}$$

acceleration = 1.96 m/s^{2}

.

Alan
Jan 4, 2015