Newton's Third Law-two masses on a wedge

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The discussion focuses on applying Newton's Third Law to a system involving two masses on a wedge. The equations derived include the tension and forces acting on both masses, leading to an expression for acceleration. Substituting specific values results in an acceleration of 0.69 m/s². The final velocity is calculated using the kinematic equation, confirming the approach taken is correct. The overall analysis appears sound, and the second part of the problem should follow logically from the first.
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The Attempt at a Solution



8 kg block:

T-m2g = -m2a

T-m1g sin(theta)- u m1g cos (theta) = m1a
T -m2g=-m2a

subtract equations:

-m1gsin(theta)-umigcos(theta) + m2g =
(m1+m2)a

or a = g(m2-m1sin(theta)-m2cos(theta)/(m1+m2)

substituting values:

a=g(8-10sin(30)-0.2(10)cos30)/(18)

a=0.69m/s/s

b)

vf^2=v0^2+2ad to find final velocity, knowing v0=0, a=0.69m/s/s and d=0.4m
 
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Your 1st part looks good so therefore i think the second part should be alright since its just subsituting values inside. So what's your question haha:biggrin:
 
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