## anonymous one year ago 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.

1. anonymous

Can I see your formulas given

2. 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

3. 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.

4. 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.

5. anonymous

Thank you :D