Solving a 2-Box Pulley System: Tension, Friction & Kinetic Energy

Finally, we can calculate the kinetic energy of the left hand box when it reaches the bottom of the plane. Using the conservation of mechanical energy, we can write:Kinitial + Uinitial = Kfinal + UfinalSince the box starts from rest, its initial kinetic energy is zero. Also, there is no change in potential energy, so Uinitial = Ufinal. Therefore, we can write:Kfinal = mgh = (10 kg)(10 m/s^2)(2 m sin 60°) = 100 JThis is the kinetic energy of the left hand box when it reaches the bottom of the plane.In summary, using the given information and equations, we have found that
  • #1
13rainboy
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Homework Statement



2 10 kilogram boxes are connected by a massless string that passes over a massless frictionless pulley. The boxes remain at rest, with the one on the right hanging vertically and the one on the left 2 meters from the bottom of an inclined plane that makes an angle of 60 degrees with the horizontal. the coefficients of kinetic friction and static friction between the left hand box and the plane are .15 and .3. you may use gravity equals 10 meters per second squared. what is the tension? determine the magnitude of the frictional force acting on the box on the plane. Now the string is cut and the left hand box slides down the inclined plane. determine the amount of mechanical energy that is converted into thermal energy during the slide to the bottom. determine the kinetic energy of the left hand box when it reaches the bottom of the plane.

Homework Equations





The Attempt at a Solution

i got 86.6 for tension, 15 Newtons for magnitude, 158.2 jewels for mechanical energy into thermal, and i got 158.2 for kinetic energy at bottom.
 
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  • #2


Hello,

Thank you for sharing your attempt at solving this problem. I can see that you have correctly used the given information and equations to find the tension, magnitude of frictional force, mechanical energy converted into thermal energy, and kinetic energy at the bottom. However, in order to provide a complete and accurate response, I would like to explain the steps and reasoning behind these calculations.

First, let's start by drawing a diagram to visualize the situation:

[Insert diagram]

As mentioned in the problem statement, the boxes are connected by a massless string and the pulley is also massless and frictionless. This means that the tension in the string will be the same on both ends. Using Newton's second law, we can write an equation for each box:

Box on the left (sliding down the inclined plane):
ΣF = ma
T - μkN = ma

Box on the right (hanging vertically):
ΣF = ma
mg - T = 0

We can solve for T in both equations:
T = μkN + ma
T = mg

Since the tension is the same in both equations, we can set them equal to each other and solve for N:
μkN + ma = mg
N = (mg - ma) / μk

Now, we can plug in the given values and solve for N:
N = (10 kg)(10 m/s^2 - 10 m/s^2 sin 60°) / (0.15)
N = 86.6 N

This is the normal force acting on the box on the inclined plane, which is also equal to the magnitude of the frictional force. Therefore, the magnitude of the frictional force is 86.6 N.

Next, we can calculate the mechanical energy converted into thermal energy during the slide to the bottom. This can be done by using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the frictional force is equal to the change in mechanical energy:

W = ΔE = ΔK + ΔU = ΔK + 0 (since there is no change in potential energy)
W = μkNΔx = μkN(2 m) = (0.15)(86.6 N)(2 m) = 25.98 J

This is the amount of mechanical energy converted
 

Related to Solving a 2-Box Pulley System: Tension, Friction & Kinetic Energy

1. What is a 2-box pulley system?

A 2-box pulley system is a simple machine that consists of two boxes connected by a rope or cable that passes over a pulley. It is used to lift or move objects by applying a force on one end of the rope.

2. How do you calculate the tension in a 2-box pulley system?

The tension in a 2-box pulley system can be calculated by using the formula T = m1g + m2g, where T is the tension, m1 and m2 are the masses of the boxes, and g is the acceleration due to gravity (9.8 m/s^2).

3. How does friction affect a 2-box pulley system?

Friction in a 2-box pulley system can reduce the tension in the rope, making it more difficult to lift or move the objects. This is because friction creates a force that acts in the opposite direction of the motion of the rope.

4. What is kinetic energy and how is it related to a 2-box pulley system?

Kinetic energy is the energy an object possesses due to its motion. In a 2-box pulley system, kinetic energy is related to the velocity of the boxes and the tension in the rope. As the boxes move, they possess kinetic energy, and the tension in the rope is responsible for maintaining their motion.

5. How can you increase the efficiency of a 2-box pulley system?

The efficiency of a 2-box pulley system can be increased by reducing friction and using lighter boxes. Reducing friction can be achieved by using smoother ropes or lubricating the pulley, while using lighter boxes will require less tension to lift or move them, resulting in less energy loss due to friction.

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