A horizontal force of 600 N pulls two masses 10Kg and 20k lying on a frictionless surface Connected by a light string.what is the tension in the string? Does the answer depend on which mass end the pull is applied?
To find the tension in the string, we need to consider the forces acting on the masses.
First, let's assume that the 10 kg mass is on the left side (closer to the force) and the 20 kg mass is on the right side (farther from the force).
When the horizontal force pulls the masses, the force will cause both masses to accelerate. The acceleration of the system can be determined by using Newton's second law: F = ma, where F is the net force, m is the total mass, and a is the acceleration.
Since the masses are connected by a light string, they will have the same acceleration. Therefore, the total mass (m) in this case will be the sum of the two masses (10 kg + 20 kg = 30 kg).
We are given that the horizontal force is 600 N. So, substituting the values into the equation F = ma, we have:
600 N = 30 kg * a
Rearranging the equation to solve for acceleration (a), we get:
a = 600 N / 30 kg
a = 20 m/s^2
Now that we have determined the acceleration of the system, we can find the tension in the string. The tension in the string is equal to the force exerted by the 20 kg mass, pulling the 10 kg mass.
We can use Newton's second law for the 20 kg mass to find the tension. The force on the 20 kg mass is its mass (20 kg) multiplied by its acceleration (20 m/s^2):
Tension = 20 kg * 20 m/s^2
Tension = 400 N
Therefore, the tension in the string is 400 N. This result is independent of which mass end the force is applied to.
Even if the forces are reversed (10 kg mass on the right and 20 kg mass on the left), the acceleration and the tension will be the same (20 m/s^2 and 400 N, respectively). This is because the magnitudes of the forces acting on both masses are still the same (600 N), even though they are in opposite directions.
So, the answer does not depend on which mass end the pull is applied. The tension in the string remains the same.
To find the tension in the string, we can start by analyzing the forces acting on each of the masses.
1. Mass 1 (10 kg): Let's denote this mass as M1. The only force acting on M1 is the tension in the string, which we'll call T1.
2. Mass 2 (20 kg): Let's denote this mass as M2. Similar to M1, the only force acting on M2 is the tension in the string, but we'll call it T2.
Since the masses are connected by a string, the tension in the string on one side is equal to the tension on the other side. Therefore, T1 = T2, regardless of which end the horizontal force is applied to.
Now, when the horizontal force of 600 N is applied, both masses will experience an acceleration. Since the surface is frictionless, there is no opposing force due to friction.
Using Newton's second law (F = m * a), we can find the acceleration of the system:
For M1, F1 = m1 * a,
where F1 = 600 N (applied force), and m1 = 10 kg (mass of M1).
Solving for a1, we find a1 = F1 / m1 = 600 N / 10 kg = 60 m/s².
For M2, F2 = m2 * a,
where F2 = 600 N (applied force), and m2 = 20 kg (mass of M2).
Solving for a2, we find a2 = F2 / m2 = 600 N / 20 kg = 30 m/s².
Now that we have the accelerations, we can find the tensions in the string.
For M1, since T1 is the only force acting on M1, we have:
T1 = m1 * a1 = 10 kg * 60 m/s² = 600 N.
For M2, the tension T2 and the applied force F2 are the only forces acting on M2, so:
F2 - T2 = m2 * a2.
Substituting the given values, the equation becomes:
600 N - T2 = 20 kg * 30 m/s².
Solving for T2, we find:
T2 = 20 kg * 30 m/s² - 600 N = 600 N.
Therefore, the tension in the string is the same regardless of which mass end the pull is applied. The tension in the string is 600 N.