Two packing crates of masses 7.93 kg and 5.13 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 5.13 kg crate lies on a smooth incline of angle 40.0°. a) Find the acceleration of the 5.13 kg crate (b) find the tension in the string.

To find the acceleration of the 5.13 kg crate, we can apply Newton's second law of motion.

Step 1: Identify the forces acting on the 5.13 kg crate.
The forces acting on the crate are the force of gravity (mg) and the tension in the string (T). Note that there is no normal force acting on the crate since it is on a smooth incline.

Step 2: Break down the forces into components.
The force of gravity can be broken down into two components: the component parallel to the incline (mg sinθ) and the component perpendicular to the incline (mg cosθ).

Step 3: Apply Newton's second law in the direction of motion.
Since the crate is on an incline, the direction of motion is along the incline. Newton's second law in this direction can be written as:
ma = mg sinθ - T

Step 4: Solve for the acceleration.
Substituting the known values, we get:
(5.13 kg) * a = (5.13 kg) * (9.8 m/s^2) * sin(40.0°) - T

Simplifying, we have:
5.13a = (5.13)(9.8)(sin(40.0°)) - T

To solve for the acceleration, we need to determine the tension in the string.

Step 5: Find the tension in the string.
To find the tension, we can consider the other crate with mass 7.93 kg. Since the two crates are connected by a light string that passes over a frictionless pulley, they are both subject to the same tension.

Step 6: Apply Newton's second law to the 7.93 kg crate.
Since the 7.93 kg crate is not on an incline, the force of gravity can be broken down into two components: the component parallel to the vertical direction (mg) and the component perpendicular to the vertical direction (mg cosθ).

Step 7: Apply Newton's second law in the vertical direction.
Applying Newton's second law in the vertical direction, we have:
(7.93 kg) * g = (7.93 kg) * (9.8 m/s^2) * cos(40.0°) - T

Simplifying, we get:
7.93g = (7.93)(9.8)(cos(40.0°)) - T

Step 8: Solve for the tension.
Since the tension in the string is the same for both crates, we can equate the expressions for T obtained from steps 4 and 8:
(5.13)(9.8)(sin(40.0°)) - T = (7.93)(9.8)(cos(40.0°)) - T

Simplifying, we have:
(5.13)(9.8)(sin(40.0°)) = (7.93)(9.8)(cos(40.0°)) - T

Solving for T, we find:
T = (7.93)(9.8)(cos(40.0°)) - (5.13)(9.8)(sin(40.0°))

Finally, substitute the value of T into the expression for acceleration obtained in step 4 to find the acceleration of the 5.13 kg crate.

To find the acceleration of the 5.13 kg crate and the tension in the string, we can analyze the forces acting on the system.

Let's break the analysis into two parts: the 7.93 kg crate and the 5.13 kg crate.

1. Analysis for the 7.93 kg crate:
- The only force acting on this crate is its weight (mg).
- The weight can be broken down into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ).
- The net force acting on this crate is the component parallel to the incline (mg*sinθ).
- Using Newton's second law (F=ma), we can equate the net force to the product of mass and acceleration:
mg*sinθ = (7.93 kg) * a1
Here, "a1" is the acceleration of the 7.93 kg crate.

2. Analysis for the 5.13 kg crate:
- The weight of the 5.13 kg crate can also be broken down into two components: one parallel to the incline (mg*sinθ) and one perpendicular to the incline (mg*cosθ).
- The net force acting on this crate is the component parallel to the incline (mg*sinθ) minus the tension in the string.
- Using Newton's second law (F=ma), we can equate the net force to the product of mass and acceleration:
mg*sinθ - T = (5.13 kg) * a2
Here, "a2" is the acceleration of the 5.13 kg crate.

Now, let's solve for the acceleration of the 5.13 kg crate and the tension in the string.

a) Solving for the acceleration of the 5.13 kg crate (a2):
- Rearranging the equation for the 5.13 kg crate, we have:
a2 = (mg*sinθ - T) / (5.13 kg)
- However, we need the tension in the string (T) in order to calculate this acceleration.

b) Solving for the tension in the string (T):
- Rearranging the equation for the 7.93 kg crate, we have:
a1 = mg*sinθ / (7.93 kg)
- Since the two crates are connected by a light string passing over a frictionless pulley, the two crates must have the same acceleration.
- Therefore, a1 = a2.
- We can equate the two acceleration equations and solve for T:
(mg*sinθ) / (7.93 kg) = (mg*sinθ - T) / (5.13 kg)
- Now, solve for T.

By solving these equations, you can find the acceleration of the 5.13 kg crate (a2) and the tension in the string (T).