a horizontal force of 600Npulls two masses 10kg lying on a frictionnnnless 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 determine the tension in the string, we need to analyze the forces acting on the masses. In this scenario, there are two masses connected by a light string and being pulled by a horizontal force.
Now, let's break down the problem into smaller parts:
1. Forces on the first mass (M1 = 10 kg):
- There is the tension force, T, pulling M1 to the right.
- There is no opposing force due to the absence of friction.
2. Forces on the second mass (M2 = 10 kg):
- There is the tension force, T, pulling M2 to the right.
- There is no opposing force due to the absence of friction.
3. Connecting the masses:
- Since the string connecting both masses is light, it has negligible mass.
- Thus, the tension force (T) is the same throughout the string.
Now, to calculate the tension in the string, we can use Newton's second law that states F = ma. In this case, the force acting on each mass is the tension (T). Therefore, we can write:
For M1:
T = M1 * a1
For M2:
T = M2 * a2
Since the masses are connected by the string, they will accelerate together. Hence, their accelerations will be equal: a1 = a2 = a.
Now, let's apply the horizontal force (F = 600 N) to the system. Since there is no friction, the net force acting on both masses is the force applied. Therefore:
F = (M1 + M2) * a
600 N = (10 kg + 10 kg) * a
600 N = 20 kg * a
To solve for 'a', divide both sides of the equation by 20 kg:
a = 600 N / 20 kg
a = 30 m/s^2
Now that we have the acceleration, we can substitute it into one of the previous equations to find T:
T = M1 * a
T = 10 kg * 30 m/s^2
T = 300 N
So, the tension in the string is 300 Newtons.
Now, to answer the second part of your question, does the answer depend on which mass end the pull is applied to?
No, the answer remains the same regardless of which mass end the pull is applied to because the tension in the string is equal throughout. The tension is determined by the total mass of the system and the applied force.