Tension, kinetic friction, and two blocks

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SUMMARY

The discussion centers on a physics problem involving two blocks connected by a rope, subjected to a horizontal force of 68 N. The first block weighs 18 kg and the second weighs 12 kg, with a coefficient of kinetic friction of 0.100. The calculated acceleration of the system is 1.3 m/s², and the tension in the rope is determined to be 95 N. The participants emphasize the importance of correctly identifying the forces acting on the blocks, including tension, normal force, weight, and kinetic friction.

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  • Understanding of Newton's laws of motion
  • Basic knowledge of forces: tension, normal force, weight, and friction
  • Ability to solve equations involving multiple variables
  • Familiarity with the concept of acceleration in physics
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Hey guys,

I have a physics problem that I tried to solve but I am not positive of the answer I got. Can someone guide me through the steps of solving it.

Problem: Two blocks connected by a rope are dragged by a horizontal force (68 N
). The first block is 18Kg, the second block is 12Kg, and the coefficient of kinetic friction is 0.100. What is the tention of the rope between the two blocks. What is the acceleration of the system.

For the acceleration, I got 1.3m/s^2, and the tention turned out to be 95 N. It's the tention that I'm really iffy about. But that could be due to a miscalculation of my acceleration.​
 
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Equations for each block

[tex]T_{applied} - T_{block} - \mu m_{2}g = m_{2}a[/tex]

[tex]T_{block} - \mu m_{1}g = m_{1}a[/tex]

Solving:

[tex]\frac{T_{applied} - \mu m_{2}g - \mu m_{1}g}{m_{1}+m_{2}} = a[/tex]

Then

[tex]T_{block} - \mu m_{1}g = m_{1}a[/tex]

Plugging
[tex]T_{block} - \mu m_{1}g = m_{1}(\frac{T_{applied} - \mu m_{2}g - \mu m_{1}g}{m_{1}+m_{2}})[/tex]
 


Hi there,

First of all, great job attempting to solve the problem on your own! It's always good to have a go at it before seeking help.

To solve this problem, we first need to identify the forces acting on the blocks. These are:

1. The tension force in the rope, which is pulling the blocks together.
2. The normal force, which is the force exerted by the surface on the blocks, perpendicular to the surface.
3. The weight of each block, which is the force pulling each block towards the center of the earth.
4. The force of kinetic friction, which is the force that opposes the motion of the blocks.

Since the blocks are connected by a rope, they will have the same acceleration. This means that we can write one equation for the net force acting on both blocks:

Net Force = (mass of first block + mass of second block) x acceleration

Now, let's break down the forces acting on the blocks and plug them into the equation:

1. Tension force: Since the rope is pulling both blocks towards the right, the tension force will also be towards the right. This means that it will contribute positively to the net force.

2. Normal force: Since the blocks are on a horizontal surface, the normal force will be equal to the weight of the blocks. This means that it will not contribute to the net force.

3. Weight of each block: The weight of each block will be directed downwards, towards the center of the earth. Since the blocks are not moving vertically, this force will also not contribute to the net force.

4. Force of kinetic friction: This force will be directed in the opposite direction of motion, which is towards the left. This means that it will contribute negatively to the net force.

Now, let's plug in the values:

Net Force = (18 kg + 12 kg) x acceleration = 30 kg x acceleration

And the forces:

68 N (tension) - 12 kg x 9.8 m/s^2 (weight of first block) - 18 kg x 9.8 m/s^2 (weight of second block) - (0.100 x 30 kg x 9.8 m/s^2) (force of kinetic friction) = 30 kg x acceleration

Simplifying this, we get:

68 N - 176.4 N - 294 N + 29.4 N = 30
 

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