- #1
Zobrox
- 5
- 0
Hi guys, really confusing myself over grasping this principle... I think my main problem is understanding datums and how choosing the direction of a positive axis effects the result.
I understand that the work energy principle states:
Wnet = ½m(vf2+vi2)
I have previously been taught that Wnet = losses - gains.
Whilst this worked for my previous studies, its now proving that I need more of a clear understanding in order to approach the question correctly.
it is given that:
Ma = 30kg
Mb = 20kg
μk = 0.1
[/B]
we are asked to work out the speed of block A, given that B has moved up the plane 0.6m.
(the poorly drawn image of the problem is below)
please note, that the string on the pulley is light and inextensible, the lengths Sa and Sb I labelled since these vary.
I have had countless of attempts... really I am looking for an approach and a way of thinking that will not only allow me to solve this problem, but countless of others like it. Meanwhile Ill try solve it some more...
My most recent attempt was to think about each object separately...
- I first differentiated the 2 varying lengths of the rope to find a relationship between the velocities... I found that 2Va = -Vb (my question here is, since they are negated how do I go about treating the reference axis for both block A and block B, since they are two differently inclined slopes)
- After which I also said for 2×ΔSa + ΔSb = 0... this then gives us a relationship of the change in distance... we know ΔSb = 0.6, therefore the other length is -0.3
- now this is where I get a bit confused... As I stated, i recently considered each object separately... then my thought process was that for any force opposing motion, then this will give us a negative work... so formulated 2 separate energy equations for block A and B...
- I thought that since block B is going up the incline, the tension force in the rope will be a positive work, and the weight and frictional forces will provide a negative work... then let this equal the final kinetic energy for B, I subbed in 2Va = -Vb however I ignored the signs, since I was just assuming a positive velocity up the slope (again bit confused here)...
- I did the same for block B, however the tension work 2T*0.3 I took away from the net work, since it's opposing block A's motion... I added the work from the weight and deducted the work from friction.
(another question is... forces opposing motion are meant to be doing negative work, this is why I added the work done by the weight on above... however gravitational potential energy is meant to decrease as you get lower in height, how does this fit in with my way of thinking with regards to the negative work?)
- So after getting the two equations I added them together, this cancels out the work by the tension, since ones positive and the others negative. However after solving all of this, the final velocity that I calculate turns out to be wrong...
Thanks to anyone who takes the time to read and answer my question(s)...
My working:
So I figured: [tex] 2S_{a} = -S_{b}, thus: 2V_{a} = -V_{b}. [/tex]
for Block A:
[tex] \mu (m_{a}g\cos{\theta_{a}})*S_{a} + 2T(S_{a}) = S_{a}\sin{\theta_{a}}*m_{a}*g - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]
for Block B:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*V_{b}^2 [/tex]
however I sub in the previous relationship [tex] 2V_{a} = -V_{b}. [/tex]
I ignore the sign however, due to me treating each system separately:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 [/tex]
then I added the two equations together... giving me:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} + \mu (m_{a}g\cos{\theta_{a}})*S_{a}= S_{a}\sin{\theta_{a}}*m_{a}*g - S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]
then I finally solve for [tex] V_{a} [/tex] and get the wrong answer :( Where am I going wrong?!
Any general input on this topic would be really appreciated... its driving me nuts.
I understand that the work energy principle states:
Wnet = ½m(vf2+vi2)
I have previously been taught that Wnet = losses - gains.
Whilst this worked for my previous studies, its now proving that I need more of a clear understanding in order to approach the question correctly.
Homework Statement
it is given that:
Ma = 30kg
Mb = 20kg
μk = 0.1
[/B]
we are asked to work out the speed of block A, given that B has moved up the plane 0.6m.
(the poorly drawn image of the problem is below)
please note, that the string on the pulley is light and inextensible, the lengths Sa and Sb I labelled since these vary.
The Attempt at a Solution
I have had countless of attempts... really I am looking for an approach and a way of thinking that will not only allow me to solve this problem, but countless of others like it. Meanwhile Ill try solve it some more...
My most recent attempt was to think about each object separately...
- I first differentiated the 2 varying lengths of the rope to find a relationship between the velocities... I found that 2Va = -Vb (my question here is, since they are negated how do I go about treating the reference axis for both block A and block B, since they are two differently inclined slopes)
- After which I also said for 2×ΔSa + ΔSb = 0... this then gives us a relationship of the change in distance... we know ΔSb = 0.6, therefore the other length is -0.3
- now this is where I get a bit confused... As I stated, i recently considered each object separately... then my thought process was that for any force opposing motion, then this will give us a negative work... so formulated 2 separate energy equations for block A and B...
- I thought that since block B is going up the incline, the tension force in the rope will be a positive work, and the weight and frictional forces will provide a negative work... then let this equal the final kinetic energy for B, I subbed in 2Va = -Vb however I ignored the signs, since I was just assuming a positive velocity up the slope (again bit confused here)...
- I did the same for block B, however the tension work 2T*0.3 I took away from the net work, since it's opposing block A's motion... I added the work from the weight and deducted the work from friction.
(another question is... forces opposing motion are meant to be doing negative work, this is why I added the work done by the weight on above... however gravitational potential energy is meant to decrease as you get lower in height, how does this fit in with my way of thinking with regards to the negative work?)
- So after getting the two equations I added them together, this cancels out the work by the tension, since ones positive and the others negative. However after solving all of this, the final velocity that I calculate turns out to be wrong...
Thanks to anyone who takes the time to read and answer my question(s)...
My working:
So I figured: [tex] 2S_{a} = -S_{b}, thus: 2V_{a} = -V_{b}. [/tex]
for Block A:
[tex] \mu (m_{a}g\cos{\theta_{a}})*S_{a} + 2T(S_{a}) = S_{a}\sin{\theta_{a}}*m_{a}*g - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]
for Block B:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*V_{b}^2 [/tex]
however I sub in the previous relationship [tex] 2V_{a} = -V_{b}. [/tex]
I ignore the sign however, due to me treating each system separately:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} - T(S_{b}) = S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 [/tex]
then I added the two equations together... giving me:
[tex] \mu (m_{b}g\cos{\theta_{b}})*S_{b} + \mu (m_{a}g\cos{\theta_{a}})*S_{a}= S_{a}\sin{\theta_{a}}*m_{a}*g - S_{b}\sin{\theta_{b}}*m_{b}*g - \frac{1}{2}*m_{b}*(2V_{a})^2 - \frac{1}{2}*m_{a}*V_{a}^2 [/tex]
then I finally solve for [tex] V_{a} [/tex] and get the wrong answer :( Where am I going wrong?!
Any general input on this topic would be really appreciated... its driving me nuts.
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