A spring has a force constant of 100N/m and an unstretched lenght of .07 m. One end is attached to a post that is free to rotate in the center of a smooth table. The other end is attatched to a 1 kg disc moving in uniform circular motion on the table, which stretches the spring by .03m. Friction is negligible. What is the centripital force on the disc?

Question

a).3 N
b)3 N
c)10 N
d)300 N
e)1,000 N

I know that F_c= (mv^2)/r, where m=1kg, v= 100 ( is this correct?), r= .07 +.03 when i pluged this in I got 100000N. What would I do with the foce constant 100N/m? this isn't the velocity. What is the correct answer?

There's also a second part to this: What is the work done on the disc by the spring during one full circle?
a)0 J
b)94 J
c)186 J
d)314 J
e)628 J

Would I use W =fd where f is the answer of the above question times (.1)?

First part; you are thinking too hard. What is the amount of force required to move the spring .03m if k= 100N/m ?

Second Part:Yes, except d and F are perpendicular, Work= force*distance SinTheta.

So for the 1st part:to find the force I did this 100/.03 but I got 333.333 I can't figure out what i am doing wrong.

Answer 1

b and a

Answer 2

give up

Answer 3

Question 1) Just use Hooke's law. F = K(change of x)*

Note: I don't how to put the little triangle thingies, so "change of" is the best I could do :/

Then plug stuff in. F = (100)(0.03) = 3 N

Question 2: Work is the change of kinetic energy. However, kinetic energy doesn't change at all, so according to W = (change of K), the work is 0J

Answer 4

what about work of spring, aka (1/2)kx^2

Answer 5

ASDF - And so x = what??? 0 - then work = 0. Still tho