Calculating Speed and Tension in a Two-Block System

[cdotter]

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:
F_gravity: (2.25kg)(9.8m/s²) = 22.05N
F_normal: 22.05N
F_friction: (22.05N)(0.450) = 9.92N
Resultant force: 22.05N - 9.92N = 12.13N

Block B:
F_gravity: (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.

 

[tiny-tim]

Hi cdotter!

Block A:
…
F_normal: 22.05N
F_friction: (22.05N)(0.450) = 9.92N
Resultant force: 22.05N - 9.92N = 12.13N

Sorry, but this is completely wrong …

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. ​

 

[cdotter]

Hi cdotter!


Sorry, but this is completely wrong …

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. ​

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?

 

[tiny-tim]

Your normal force and frictional force were correct, but your "resultant force" equation was completely wrong.

Try again, using T. ​

 

[cdotter]

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.

 

[tiny-tim]

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.

 

[cdotter]

No, everything is accelerating, and you've left "a" out of your F = ma.

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

Block A: $F_{fric}-T_{rope}=ma$
Block B: $F_{gravity}-T_{rope}=ma$

Right?

edit: I guess not. I'm so lost haha.

 

[tiny-tim]

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

Start with the lower block …

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

 

[cdotter]

Start with the lower block …

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

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.

 

[cdotter]

I figured it out by reading another post, thank you for the hints tiny-tim.