A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.10,find the tension in the string.
A) 19 N
B) 24 N
C) 35 N
D) 38 N

Question

A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.10,find the tension in the string. 
A) 19 N 
B) 24 N 
C) 35 N 
D) 38 N

Quizplus · Accepted Answer

Since the two masses are connected by a string over a pulley, they will accelerate together with the same acceleration. We can use Newton's second law to set up equations for the forces acting on each object in the vertical and horizontal direction. In the vertical direction, the weight of the hanging mass and the tension in the string balance out to give a net force of 0. In the horizontal direction, the force of friction opposes the tension in the string, giving a net force of ma, where a is the acceleration of the system.

For the hanging weight: 
T - mg = 0 
T = mg = (9.0 kg)(9.8 m/s²) = 88.2 N

For the sliding block: 
μkN - T = ma 
μkmg - T = ma

We can eliminate the normal force N since it is equal to the weight of the block: 
μkmg - T = ma 
(0.10)(5.0 kg)(9.8 m/s²) - T = (14.0 kg)a

We also know that the acceleration of the system is the same for both objects, so we can equate the expressions for a: 
a = T/(5.0 kg) = (μkmg - T)/(14.0 kg) 
Solving for T: 
T = 35 N

A 90-Kg Hanging Weight Is Connected by a String Over a a Pulley