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anonymous
 5 years ago
Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 7.00 kg crate lies on a smooth incline of angle 41.0°. Find the acceleration of the 7.00 kg crate.
anonymous
 5 years ago
Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley as in the figure below. The 7.00 kg crate lies on a smooth incline of angle 41.0°. Find the acceleration of the 7.00 kg crate.

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Owlfred
 5 years ago
Best ResponseYou've already chosen the best response.0Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the equations of the motion : where T is the tension in the rope, the system will have common acceleration a since the same rope. 10g T = 10a..........(1) T 7g =7a...............(2) Solving both the equations, a= 3g/17. ( g= 9.8m/s^2) a= 1.72m/s^2 hope this will help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[Let \]'T' be the tension of the string and 'a' be the acceleration. Consider the motion of the mass m1=10 kg, According to newton's second law, W1T=m1a m1gT=m1a (W=mg) _______(1) Now consider the motion of body of mass m2=7 kg, According to newton's second law, TW2sin41=m2a Tm2gsin41=m2a _______(2) (1)+(2)= m1gT+Tm2gsin41=ma+m2a m1gm2gsin41=m1a+m2a g(m1m2sin41)=a(m1+m2) a=g(m1m2sin41)/(m1+m2) a=9.81(10=7*sin41)/(10+7) a=3.12m/s^2 Therefore acceleration of the body of mass 7 kg = 3.12 m/s^2
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