anonymous
  • anonymous
A 3.2-kg sphere is suspended by a cord that passes over a 1.8-kg pulley of radius 3.6 cm. The cord is attached to a spring whose force constant is k = 86 N/m. Assume the pulley is a solid disk. (a) If the sphere is released from rest with the spring unstretched, what distance does the sphere fall through before stopping? (b) Find the speed of the sphere after it has fallen 25 cm.
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Can I see your formulas given
anonymous
  • anonymous
Idk, but this looks right A.)f=ma=f=kx add both masses=5kg f=5*9.9= 49N 49=86x x=.5697m EPE=1/2*86*.5607^2 =13.95 Joules PE=KE+=EPE 13.95=5*9.8*h h=.2847m B.) PE=KE=EPE 13.95=1/2*5*v^2 PE=5*9.8*25 =1225Joules 1225=1/2*5*V^2 v=22m/s
anonymous
  • anonymous
@n648c788 Not quite but good try :) You're method for part A would work well for a statics problem, unfortunately, we have to consider the motion as well. The number you have found would be the point at which the forces on the sphere balance, but that just means zero acceleration. The mass is still moving with a velocity downwards so it still travels further before the mass stops Honestly I'd solve this with a simple differential equation, but I'm not sure if that exceeds the scope of the question.

Looking for something else?

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

More answers

IrishBoy123
  • IrishBoy123
when the sphere finally stops, its grav potential will be converted to spring potential. if it falls distance x, then \(mgx = \frac{1}{2} k x^2\), as the spring will have extended x. the pulley and the sphere will be at rest at this point. as for "speed of the sphere after it has fallen 25 cm", at that point you will have to consider grav and spring potential energy BUT the sphere and the disc/pully will also be moving. they will have kinetic energies given by \(\frac{1}{2}m v^2 + \frac{1}{2}I \omega^2\) where \(v = 0.036 \omega\) , using the radius of the disc/pully. thus \(mg(0.25) = \frac{1}{2} k (0.25)^2 + \frac{1}{2}m v^2 + \frac{1}{2}I \omega^2\), where \(v = 0.036 \omega\), and Moment of Inertia for a disc: \( I = \frac{1}{2}M R^2\) this is an oscillating system and you could build a D.E, but seems OTT given the nature of the questions.
anonymous
  • anonymous
Thank you :D

Looking for something else?

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