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Old Jun 11th 2017, 02:09 AM   #1
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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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Old Jun 11th 2017, 04:48 AM   #2
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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.
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Old Jun 11th 2017, 06:55 AM   #3
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conservation of energy ...

$Mgx + \dfrac{1}{2}kx^2 = mgh + \dfrac{1}{2}(M+m)v^2$

$v = \sqrt{\dfrac{2g(Mx-mh)+kx^2}{M+m}}$

$x = 0.2 \, m$, $h = 0.2\sin(40^\circ) \, m$, $M=30 \, kg$, $m=20 \, kg$, and $k = 250 \, N/m$
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Old Jun 11th 2017, 07:27 AM   #4
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Dear friends Skeeter

Thank you soooooooo much .
Now , the question has solved .

Best regards
Razi
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