# Distance After Release- Problem on Forces

1. Oct 30, 2009

### haengbon

1. The problem statement, all variables and given/known data

In figure A, the coefficient of friction between block A and the tabletop is 0.3, Mass of block A= 30 kg, Mass of block B=20 kg. How far will block B drop in the first 3 s after the system is released from rest?

here is the figure :D

http://img137.imageshack.us/img137/5771/imagenz.png [Broken]

2. Relevant equations

....

3. The attempt at a solution

*we use gravity= 10 (not 9.8 )

w (weight)
Nf (normal force)
ΣF (semation of forces)
f (friction)
T (tension)

BLOCK A

ΣFy= 0

-w + NF = 0
Therefore w = NF

W= mg
= 30 (10)
= 300 N

Therefore NF = 300 N

f= μNF
= 0.3 (300 N)
f= 90 N

ΣFx= ma

-f + T = ma
-90 + T = ma
-90 + T= 30 (a)
T= 90 + 30a

BLOCK B

W=mg
W=20 (10)
W= 200 N

ΣFy= ma

T-w=ma
T- 200 N = 20a
T= 200+20a

T=T

90+30a = 200 + 20a
90- 200 = 20a – 30a
-110 = -10a
------- -------
-10 -10
a= 11 m/s^2

y= Vo + Voyt + at^2/2
y= 0 + 0 + 11 (3) 2 /2
y = 49.5 m

I got 49.5 but my professor said it was wrong :(can someone tell me where I went wrong? and what IS the correct answer?

Last edited by a moderator: May 4, 2017
2. Oct 30, 2009

### Staff: Mentor

You made a sign error in setting up your Fy equation. You need to be consistent in your use of "a" for acceleration. If mass A moves to the right, mass B moves down. Thus if the acceleration of mass A is "a", then the acceleration of mass B is "-a".

Note how your acceleration is greater than gravity! That can't be right.

Just fix that equation for block B above and you'll be fine.

3. Oct 30, 2009

### haengbon

Oh so the only mistake I made was not making the acceleration of block B negative? :D so if I changed it to negative, then it'll make all the difference ^^? I asked through yahoo answers too and one replied that the answer was 9.7 m :D is it most likely correct? I couldn't understand his explanation that well that's why I'm still confused -__-

4. Oct 30, 2009

### Staff: Mentor

Yes. Try it and see. Correct the equation as I indicated; find the correct acceleration; then use it to find the distance traveled.

5. Oct 30, 2009

### haengbon

So I tried doing it like this...

T-w=m(-a)
T- 200 N = 20(-a)
T= 200+20(-a)

Is this correct?

6. Oct 30, 2009

### Staff: Mentor

So far, so good.

7. Oct 30, 2009

### haengbon

So this is the result right? :D

T=T

90+30a = 200 + 20(-a)
90- 200 = 20(-a) – 30a

so from what I can understand... 20(-a) becomes -20a right ? :D or is it wrong?

so...
90- 200 = -20a – 30a
-110 = -50a
---------------
-50 -50

a= 2.2 m/s^2 ? :D

8. Oct 30, 2009

### Staff: Mentor

Looks good to me.

9. Oct 30, 2009

### haengbon

thank you thank you thank you so much!! I got the correct answer!!! xDD thank you for being so patient :)