First, we draw the free body diagrams and set up the equations for the problem. Let's denote the tension in the string T, the acceleration of the system a, the force of gravity on the hanging object FG, the force of kinetic friction Fk, and the weight of the block on the table W.
- For the 5.00-kg block on the table:
- The force pulling it horizontally is T.
- The force of kinetic friction Fk is acting in the opposite direction of the movement.
- The weight is W = m * g, where m = 5.00 kg and g = 9.81 m/s²
- The normal force Fn acts vertically upward and is equal to the weight W for the system to be in equilibrium in the vertical direction.
The equation for the horizontal direction can be written as:
m1 * a = T - Fk (1)
- For the 8.90-kg hanging object:
- The force of gravity FG = m * g, where m = 8.90 kg and g = 9.81 m/s²
- The tension T acts upward.
The equation for the vertical direction can be written as:
m2 * a = FG - T (2)
Now, let's calculate the force of gravity FG and the force of kinetic friction Fk.
FG = m2 * g = 8.90 kg * 9.81 m/s² = 87.309 N.
Fk = µ * Fn = µ * W = µ * m1 * g = 0.150 * 5.00 kg * 9.81 m/s² = 7.3575 N.
Next, we need to find the acceleration a of the system to solve for T.
We can eliminate T by summing equations (1) and (2):
m1 * a + m2 * a = FG - Fk
a = (FG - Fk) / (m1 + m2) = (87.309 N - 7.3575 N) / (5.00 kg + 8.90 kg) = 79.9515 N / 13.90 kg = 5.7509 m/s².
Finally, we can use this to find the tension T, either by plugging into equation (1) or (2). Let's use equation (1):
T = m1 * a + Fk = 5.00 kg * 5.7509 m/s² + 7.3575 N = 28.7545 N + 7.3575 N = 36.112 N.
The tension in the string is 36.112 N.
A 8.90-kg hanging object is connected by a light, inextensible cord over a light, frictionless pulley to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as 0.150, find the tension in the string.
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