anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Owlfred
  • Owlfred
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!
anonymous
  • anonymous
where's the diagram?
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[Let \]'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
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.