A 8.74-kg hanging weight m2 is connected by a string over a pulley to a 5.00-kg block m1 that is sliding on a flat table. If the coefficient of kinetic friction is 0.190, find the tension in the string.
To find the tension in the string, we need to consider the forces acting on both masses and apply Newton's laws of motion.
For the hanging weight (m2):
1. The weight of m2 is acting downward and can be calculated as W2 = m2 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
W2 = 8.74 kg * 9.8 m/s^2 = 85.652 N (newtons)
2. The tension in the string is acting upward.
For the block on the table (m1):
1. The weight of m1 is acting downward and can be calculated as W1 = m1 * g.
W1 = 5.0 kg * 9.8 m/s^2 = 49.0 N
2. The frictional force on the block acts opposite to the direction of its motion. The force of kinetic friction can be calculated as fk = u * N, where u is the coefficient of kinetic friction and N is the normal force.
To find the normal force, first, we need to calculate the gravitational force acting on the block perpendicular to the table, which is equal to W1. Since the block is not accelerating vertically, the normal force is equal to the gravitational force.
Therefore, the frictional force can be calculated as fk = 0.190 * W1 = 0.190 * 49.0 N = 9.31 N.
Now, let's apply Newton's second law for each mass:
For m2:
The net force is given by the difference between the tension in the string (T) and the weight (W2):
T - W2 = m2 * a, where a is the acceleration of the system.
Since m2 is hanging and is not accelerating vertically (because the string is ideal), the net force is zero. Thus, T = W2.
T = 85.652 N.
For m1:
The net force is given by the difference between the tension in the string (T) and the frictional force (fk):
T - fk = m1 * a, where a is the acceleration of the system.
The acceleration of the system can be determined by setting the force equations equal to each other:
T - fk = m1 * a = 49.0 N - 9.31 N = (m1 + m2) * a.
Substituting the given values into the equation, we get:
85.652 N - 9.31 N = (5.0 kg + 8.74 kg) * a,
76.342 N = 13.74 kg * a,
a = 76.342 N / 13.74 kg ≈ 5.55 m/s^2.
Now that we have the acceleration of the system, we can substitute it back into one of the force equations to find the tension:
T = m1 * a + fk,
T = 5.0 kg * 5.55 m/s^2 + 9.31 N,
T ≈ 27.75 N + 9.31 N,
T ≈ 37.06 N.
Therefore, the tension in the string is approximately 37.06 N.
To find the tension in the string, we need to consider the forces acting on both masses.
Let's start by analyzing the forces acting on the 8.74 kg hanging weight (m2):
1. The force of gravity acting on m2 is given by Fg2 = m2 * g, where m2 is the mass of the hanging weight and g is the acceleration due to gravity (approximately 9.8 m/s^2). So, Fg2 = 8.74 kg * 9.8 m/s^2.
2. The tension in the string pulls m2 upwards. Let's call this tension T.
Next, let's analyze the forces acting on the 5.00 kg block (m1) on the flat table:
1. The force of gravity acting on m1 is given by Fg1 = m1 * g, where m1 is the mass of the block and g is the acceleration due to gravity.
2. The frictional force acting on m1 opposes its motion on the table and is given by Ff = μ * Fn, where μ is the coefficient of kinetic friction and Fn is the normal force acting on the block. The normal force is the force exerted by the table perpendicular to the surface, which is equal to the force of gravity acting on the block (Fn = Fg1).
Now, we can use Newton's second law of motion to determine the acceleration of the system. Since both masses are connected by a string, they will have the same acceleration (a). The total force acting on m2 is T - Fg2 (since the force of gravity acts downwards), and the total force acting on m1 is Ff - T (since the tension force acts downwards).
Applying Newton's second law in the direction of motion for both masses:
For m2: T - Fg2 = m2 * a
For m1: Ff - T = m1 * a
Substituting the respective equations for Fg2 and Ff, we get:
T - (8.74 kg * 9.8 m/s^2) = 8.74 kg * a
(0.190 * (5.00 kg * 9.8 m/s^2)) - T = 5.00 kg * a
Simplifying the equations, we have:
T - 85.652 = 8.74a
0.190 * 49 - T = 5.00a
Now, we have a system of two equations and two variables (T and a). We can solve this system using algebraic manipulation or substitution methods:
Rearrange the first equation to solve for T:
T = 8.74a + 85.652
Substitute this expression for T in the second equation:
0.190 * 49 - (8.74a + 85.652) = 5.00a
Now, solve for a:
0.190 * 49 - 85.652 - 5.00a = 8.74a
0.190 * 49 - 85.652 = 13.74a
9.31 - 85.652 = 13.74a
-76.342 = 13.74a
a = -76.342 / 13.74
Once you have the value of acceleration (a), substitute it back into one of the equations to find the tension force (T):
T = 8.74 * a + 85.652
Now substitute the calculated value of a to get the tension force in the string.