an object of mass m1= 311 g on an inclined surface. The angle of the inclined surface is θ = 40o with the horizontal. The object m1 is connected to a second object of mass m2 = 355 g on the adjacent horizontal surface. Further, an external force of magnitude ІFextІ = 3.6 N is exerted on the object of mass m1. We observe both objects to accelerate. Assuming that the surfaces and the pulley are frictionless, and the pulley and the connecting string are massless, what is the tension in the string connecting the two objects?

The pulley is at the bottom of incline

It is not clear to me which way the incline is, and which direction the external force is acting.

Two masses are connected by a light string passing over a light, frictionless pulley as in Figure P5.52. The 5.00 kg mass is released from rest at a point 4.00 m above the floor.

Figure P5.52
(a) Determine the speed of each mass when the two pass each other.


m/s
(b) Determine the speed of each mass at the moment the 5.00 kg mass hits the floor.


m/s
(c) How much higher does the 3.00 kg mass travel after the 5.00 kg mass hits the floor?
m

To find the tension in the string connecting the two objects, we first need to analyze the forces acting on each object.

For the object with mass m1 on the inclined surface, there are two forces to consider. The force of gravity acting on the object is given by F_gravity1 = m1 * g, where g is the acceleration due to gravity. The other force is the external force applied, which we'll call F_ext. In this case, F_ext = 3.6 N.

For the object with mass m2 on the horizontal surface, the only force acting on it is the force of gravity, which is given by F_gravity2 = m2 * g.

Now, let's consider the tension in the string connecting the objects. Since the pulley is at the bottom of the incline, the tension in the string will be the same on both sides of the pulley. We'll call this tension T.

Since the surfaces are frictionless, the tension force T will be equal to the force required to cause the acceleration of both objects. To find this force, we'll use Newton's second law, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration: F_net = m * a.

For the object with mass m1, the net force can be expressed as F_ext - F_gravity1 = m1 * a, where a is the acceleration of both objects.

For the object with mass m2, the net force is given by F_gravity2 = m2 * a.

Next, we'll relate the acceleration of both objects to find the tension T. Since the objects are connected by a string passing over a pulley, their accelerations must be related by the equation a2 = -a1, where a1 is the acceleration of object m1. This is because as one object accelerates downwards, the other object accelerates upwards, and the magnitudes of their accelerations are equal.

Now, we can set up a system of equations:

(F_ext - F_gravity1) = m1 * a1 --> Equation 1
F_gravity2 = m2 * (-a1) --> Equation 2

Next, we can substitute F_gravity1 = m1 * g and F_gravity2 = m2 * g into the equations:

(F_ext - m1 * g) = m1 * a1 --> Equation 1
m2 * g = -m2 * a1 --> Equation 2

Now, we can solve this system of equations for the acceleration a1:

F_ext - m1 * g = m1 * a1
m2 * g = -m2 * a1

Simplifying Equation 2, we get:
g = -a1

Substituting this into Equation 1, we get:
F_ext - m1 * g = m1 * a1

Finally, we can solve for a1:
F_ext - m1 * g = m1 * a1
3.6 N - 0.311 kg * 9.8 m/s^2 = 0.311 kg * a1

Solving for a1, we find:
a1 = (3.6 N - 0.311 kg * 9.8 m/s^2) / 0.311 kg

Now, we can substitute the value of a1 into Equation 2 to find the tension T:

m2 * g = -m2 * a1
Tension = m2 * (-a1)

Substituting the given values, we get:
Tension = 0.355 kg * (-(3.6 N - 0.311 kg * 9.8 m/s^2) / 0.311 kg)