A system has three masses, a 10kg mass is lying on a level surface and is connected on the left to a 6kg mass by a cord passing over a massless,frictionless pulley.And on the right, it is connected to a 9kg mass by a cord also passing over a frictionless,massless pulley. The coefficient of kinetic friction on the 10 kg mass is 0.20.what is the acceralation of the system and the tensions in the left and right cord?
0.392m/s^2
To determine the acceleration of the system and the tensions in the left and right cords, we can use Newton's second law of motion and apply it to each of the masses.
First, let's calculate the net force acting on each mass:
For the 10kg mass:
The force of gravity acting on it is given by: F_g = m * g
Where m is the mass (10kg) and g is the acceleration due to gravity (9.8 m/s^2).
Therefore, F_g = (10 kg) * (9.8 m/s^2) = 98 N
The force of friction acting on it is given by: F_f = μ * N
Where μ is the coefficient of kinetic friction (0.20) and N is the normal force.
Since the mass is on a level surface, the normal force is equal to the force of gravity (N = F_g).
Therefore, F_f = (0.20) * (98 N) = 19.6 N
The net force acting on the 10kg mass is given by: F_net = F_applied - F_f
Since there are no externally applied forces, F_applied is zero.
Therefore, F_net = - F_f = -19.6 N (negative sign indicates opposite direction of motion)
Now, let's consider the forces acting on the 6kg mass:
The force of gravity acting on it is given by: F_g = m * g = (6 kg) * (9.8 m/s^2) = 58.8 N
Since the cords are massless and frictionless, the tension in the left cord (connected to the 6kg mass) is the same as the tension in the right cord.
Therefore, let's call the tension in both cords T.
The net force acting on the 6kg mass is given by: F_net = T - F_g
Therefore, T = F_net + F_g
Plugging in the values, T = (19.6 N) + (58.8 N) = 78.4 N
Finally, let's consider the forces acting on the 9kg mass:
The force of gravity acting on it is given by: F_g = m * g = (9 kg) * (9.8 m/s^2) = 88.2 N
The net force acting on the 9kg mass is given by: F_net = T - F_g
Therefore, T = F_net + F_g
Plugging in the values, T = (88.2 N) + (78.4 N) = 166.6 N
Since the masses are connected by a pulley, the acceleration of the system will be the same for both masses. Let's call it a.
Using Newton's second law of motion, we have:
For the 6kg mass: F_net = m * a
Therefore, T - F_g = (6 kg) * a
For the 9kg mass: F_net = m * a
Therefore, T - F_g = (9 kg) * a
Since the tensions T are the same in both cords, we can equate the two equations:
(6 kg) * a = (9 kg) * a
6a = 9a
Simplifying, we find a = 0
Therefore, the acceleration of the system is 0 m/s^2, meaning that the masses are at rest.
The tensions in the left and right cords are:
Tension in the left cord (connected to the 6kg mass): T = 78.4 N
Tension in the right cord (connected to the 9kg mass): T = 166.6 N