Calculating Speed and Tension in a Two-Block System

In summary, the conversation is about a physics problem involving two blocks connected by a cord over a pulley. The goal is to find the speed of each block after moving a certain distance and the tension of the cord. The correct approach involves using the equations F = ma and solving for the acceleration of each block, then using the kinematic equation to find their final speeds. The tension of the cord can be found by considering the forces acting on each block and their resulting acceleration.
  • #1
cdotter
305
0

Homework Statement


Block A (mass 2.25 kg) rests on a tabletop. It is connected by a horizontal cord passing over a frictionless pulley to a hanging block B (mass 1.30 kg.) The coefficient of kinetic friction between block A and the table is 0.450. After the blocks are released from rest, find a. the speed of each block after moving 3.00 cm and b. the tension of the cord.


Homework Equations





The Attempt at a Solution


Block A:
Fgravity: (2.25kg)(9.8m/s2) = 22.05N
Fnormal: 22.05N
Ffriction: (22.05N)(0.450) = 9.92N
Resultant force: 22.05N - 9.92N = 12.13N

Block B:
Fgravity: (1.30kg)(9.8m/s2) = 12.74N

I have no idea what to do after this. I found the exact same problem through a search but I don't understand the explanation.
 
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  • #2
Hi cdotter! :smile:
cdotter said:
Block A:

Fnormal: 22.05N
Ffriction: (22.05N)(0.450) = 9.92N
Resultant force: 22.05N - 9.92N = 12.13N

Sorry, but this is completely wrong :redface:

you haven't used the tension, T, at all;

and the 22.05 and the 9.92 are perpendicular, so you can't add (or subtract) them anyway.

Try again. :smile:
 
  • #3
tiny-tim said:
Hi cdotter! :smile:


Sorry, but this is completely wrong :redface:

you haven't used the tension, T, at all;

and the 22.05 and the 9.92 are perpendicular, so you can't add (or subtract) them anyway.

Try again. :smile:

I now understand that the forces are perpendicular and that I can't subtract them, but how are my normal force and frictional force calculations incorrect?
 
  • #4
Your normal force and frictional force were correct, but your "resultant force" equation was completely wrong.

Try again, using T. :smile:
 
  • #5
Block B pulls down with 12.74N of force. Block A has a frictional force of 9.92N. Shouldn't the tension then be 12.74N+9.92N = 22.66N? My book says it's 11.7N. I have absolutely no idea where 11.7N comes from.
 
  • #6
cdotter said:
Block B pulls down with 12.74N of force. Block A has a frictional force of 9.92N. Shouldn't the tension then be 12.74N+9.92N = 22.66N?

No, everything is accelerating, and you've left "a" out of your F = ma. :wink:
 
  • #7
tiny-tim said:
No, everything is accelerating, and you've left "a" out of your F = ma. :wink:

Oh, but I don't understand how to put what I have into the equation? :confused:

Block A: [itex]F_{fric}-T_{rope}=ma[/itex]
Block B: [itex]F_{gravity}-T_{rope}=ma[/itex]

Right?

edit: I guess not. I'm so lost haha.
 
Last edited:
  • #8
cdotter said:
Oh, but I don't understand how to put what I have into the equation? :confused:

Start with the lower block …

there are only two forces on it, and their total has to equal its mass times its acceleration. :smile:
 
  • #9
tiny-tim said:
Start with the lower block …

there are only two forces on it, and their total has to equal its mass times its acceleration. :smile:

Could you show me what you mean because I have absolutely no idea what to do. It has two forces acting on it: weight pulling down and the tension of the rope pulling up. I don't know how to calculate those or put them into the equation.
 
  • #10
I figured it out by reading another post, thank you for the hints tiny-tim.
 

1. What is the concept of "two blocks on a tabletop"?

The concept of "two blocks on a tabletop" refers to a common physics problem where two blocks of different masses are placed on a flat surface and interact with each other through various forces.

2. What are the forces acting on the blocks in this scenario?

The forces acting on the blocks can include gravity, normal force, and friction. These forces can vary depending on the specific situation and can affect the motion of the blocks.

3. How does the mass of the blocks affect their motion?

The mass of the blocks can affect their motion in various ways. For example, a heavier block may have a greater gravitational force and therefore be more difficult to move, while a lighter block may be easier to move. The mass can also affect the normal force and friction between the blocks and the tabletop.

4. What is the role of friction in this scenario?

Friction plays a crucial role in the motion of the blocks on the tabletop. It is the force that opposes the motion of the blocks and can be affected by factors such as the surface of the tabletop and the mass of the blocks. Friction can also cause the blocks to come to a stop if there is no external force acting on them.

5. How can the motion of the blocks be calculated in this scenario?

The motion of the blocks can be calculated using Newton's laws of motion and the equations for forces such as gravity and friction. The specific calculations will depend on the given conditions of the scenario, such as the masses and forces acting on the blocks.

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