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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.

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  1. Owlfred
    • 5 years ago
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    Hoot! 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!

  2. anonymous
    • 5 years ago
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    where's the diagram?

  3. anonymous
    • 5 years ago
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    the 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

  4. anonymous
    • 5 years ago
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    \[Let \]'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, W1-T=m1a m1g-T=m1a (W=mg) _______(1) Now consider the motion of body of mass m2=7 kg, According to newton's second law, T-W2sin41=m2a T-m2gsin41=m2a _______(2) (1)+(2)= m1g-T+T-m2gsin41=ma+m2a m1g-m2gsin41=m1a+m2a g(m1-m2sin41)=a(m1+m2) a=g(m1-m2sin41)/(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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