Two Hanging mass's attached to one on the table

[charan1]

Homework Statement

The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.

What is the acceleration of the 2.0 kg block?

Homework Equations

F=uK*Force Normal
Fnet=ma

The Attempt at a Solution

I tried to take the mass's of both the hanging and used them to calculate a net force on the block in the center and then solved for acceleration that way. Here's my work.

3kg*9.8=29.4N Right Side Tension
1kg*9.8=9.8N Left Side Tension
2kg*9.8=19.6N Force Normal

F=.35*19.6N
6.86N

(29.4N-6.86N)-9.8N=2kg*acceleration

a=6.34m/s^2

and it turns out this is wrong please help me out thank you!

-Charan

 

[nasu]

You forgot the left side tension...

 

[baba89]

Hello Charan,

Your attempt at solving for the acceleration is on the right track, but there are a few errors in your calculations. First, the force of kinetic friction (Fk) should be calculated using the coefficient of kinetic friction (uk) and the normal force (Fn), not the force of gravity (mg). So, the correct calculation for Fk would be:

Fk = uk * Fn = 0.350 * 19.6N = 6.86N

Next, when calculating the net force on the block, you need to take into account the direction of the forces. In this case, the right side tension (29.4N) and the force of kinetic friction (6.86N) are acting in the opposite direction of the left side tension (9.8N). So, the net force would be:

Fnet = (29.4N - 6.86N) - 9.8N = 12.74N

Finally, using Newton's second law (Fnet = ma), we can solve for the acceleration (a):

a = Fnet/m = 12.74N/2kg = 6.37m/s^2

Your calculation was close, but it seems that you made a small mistake in the direction of the forces. Remember to always pay attention to the direction of forces when calculating net force and acceleration. I hope this helps!