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Jun 11th 2017, 03:09 AM
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Razi
Junior Member
Join Date: Apr 2017
Location: Turkey
Posts: 15
Question No. 2
Hello again my friends
Today i have another question , could any one help me to find the solution ? i will thankful for him .
The question in the photo attached
answer : 1.25 m / sec
Thanks in advance and best regards
Razi
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Jun 11th 2017, 05:48 AM
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2
HallsofIvy
Senior Member
Join Date: Aug 2010
Posts: 240
Have you calculated the components of the force on the mass on the inclined plane parallel to and perpendicular to the incline? The component parallel to the incline is equal to the vertical force on the other mass.
Jun 11th 2017, 07:55 AM
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skeeter
Senior Member
Join Date: Aug 2008
Posts: 110
conservation of energy ...
$Mgx + \dfrac{1}{2}kx^2 = mgh + \dfrac{1}{2}(M+m)v^2$
$v = \sqrt{\dfrac{2g(Mxmh)+kx^2}{M+m}}$
$x = 0.2 \, m$, $h = 0.2\sin(40^\circ) \, m$, $M=30 \, kg$, $m=20 \, kg$, and $k = 250 \, N/m$
Jun 11th 2017, 08:27 AM
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4
Razi
Junior Member
Join Date: Apr 2017
Location: Turkey
Posts: 15
Dear friends Skeeter
Thank you soooooooo much .
Now , the question has solved .
Best regards
Razi
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