Work problem. Please see attached. I'm having trouble with part (b)

- anonymous

Work problem. Please see attached. I'm having trouble with part (b)

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- anonymous

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- anonymous

I struggled with this type of problem when I was in physics last quarter :-(

- anonymous

did you find the acceleration... since they're roped together you assume that the acceleration of each is the same

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## More answers

- anonymous

if ou have acceleration you'll have two unknowns that can be related to each other

- anonymous

Yes, I tried finding it. I got a=4.70 m/s^2

- anonymous

So for A, I did sum of the forces in x (x-axis being parallel to the inclined plane) and then for B, I did sum of the forces in y. I made sure to make the direction of motion positive. For A, there's a +T in the equation and B, there's a -T in the equation. So I added both equations so that T would cancel out and I could solve for a.

- anonymous

|dw:1359083523070:dw|

- anonymous

|dw:1359083623616:dw|

- anonymous

|dw:1359083742313:dw|

- anonymous

I just made my x-axis parallel to the plane. So my equation was m1(a) = (-mew_k)(m1)(g)(cos(40)) +T
And then for block B, I did m2(a) = (m2)(g) - T
And then I added both equations.
a(m1+m2) = (-mew_k)(m1)(g)(cos(40)) + (m2)(g)
a = (-mew_k)(m1)(g)(cos(40)) + (m2)(g)/(m1+m2)

- anonymous

\[sin(40)=\frac{W_x}{W}\]
\[W_x=mgsin(40)\]
\[W_y=mgcos(40)\]
\[\sum F_y=N-W\]
\[N=W\]
\[\sum F_{x1}=m_1a_{1x}=T-W_x-f_k=T-mgsin(40)-(\mu_s)mgcos(40)\]
\

- anonymous

where is your weight x component?

- anonymous

ha, maybe that's the problem!

- anonymous

that must've been where I goofed :-p. Just silly mistakes I guess.

- anonymous

\[m_1a_{1x}=T-m_1gsin(40)-\mu_s(m_1gcos(40))\]

- anonymous

I didn't bother to label the acceleration since it will be the same. Would that be correct?

- anonymous

\[m_2a_{2y}=T-m_2g
\]
\[a_{2y}=-a_{1x}\]

- anonymous

is the pulley frictionless?

- anonymous

yes it is

- anonymous

\[T=-m_2a_{1x}-m_2g=-m_2(a_{1x}+g)\]
\[T=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

- anonymous

Now that I added in the x-component of weight for Block A, my acceleration is 1.754 m/s^2

- anonymous

\[-m_2a_{1x}-m_2g=m_1a_{1x}+m_1gsin(40)+\mu_km_1gcos(40)\]

- anonymous

\[-m_2a_{1x}-m_1a_{1x}=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\]
\[a_{1x}(-m_2-m_1)=m_2g+m_1gsin(40)+\mu_km_1gcos(40)\]
\[a_{1x}=

- anonymous

i think i dropped a sign somewhere let me get a paper and pen

- anonymous

What was your acceleration?
m1(a) = (-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) +T
m2(a) = m2(a) = (m2)(g) - T
a(m1+m2) = (-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) + (m2)(g)
a = ((-mew_k)(m1)(g)(cos(40)) -(m1)gsin(40) + (m2)(g))/(m1+m2)

- anonymous

I have a pretty good idea on the setup of the equations, so you can just tell me if I completely missed something. And also, what your acceleration is.

- anonymous

how'd you define your axis for the y part?

- anonymous

my axis for the y-part is regular. Vertical for y, horizontal for x.

- anonymous

I mean for block B, that is.

- anonymous

isn't downwards negative for the mg?

- anonymous

When I took physics, my instructor told us to make the positive direction for each object the same as the direction of motion.

- anonymous

so for block A, up and to the right would be positive. And for block B, down would be positive.

- anonymous

so you're accounting for the system as a whole.

- anonymous

i got -1.67

- anonymous

that's for B it could be rounding error though

- anonymous

so for mine
\[a_{2x}=-1.67\]
\[a_{1x}=-a_{2x}=-(-1.67)=1.67\]

- anonymous

if you didn't round use your answer... min was rounded.... a lot . my graphing calc is out in the car and it's freezing lol. Funny thing, i just got back from physics 2

- anonymous

lol

- anonymous

We're working out of Hibbeler's book for dynamics and the chapter on work is confusing. I mean, I was already confused with these types of problems in physics I but now I just feel even more lost. This book doesn't explain things very well. You really have to read and re-read stuff in there.

- anonymous

so anyways now you have
\[S=\frac{1}{2}(1.754)t^2+vt+S_0\]
and
\[V=at+v_0=1.754t+0\]

- anonymous

so you can just sub in =]. Never heard of hibbeler. however this book from about 10 years ago im using is confusing. It uses greek lettering for everything and my teacher uses easy things like
\[R=\frac{l}{a}\]

- anonymous

my friend sent me her stuff, so I'll try it out and tell you if it worked.

- anonymous

Thank you for taking the time to help me.

- anonymous

no problem if you need anymore help let me know

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