Question

Dear Cramster Friends, I have solved this one question on my ownand I just want to confirm my answer before I submit it. Thank-youfor your help and efforts.

(Q) Albertine finds herself in a very odd contraption. She sits ina reclining chair, in front of a large, compressed spring. Thespring is compressed 5.00 from its equilibrium position, and aglass sits 19.8 from her outstretched foot.

For what value of the spring constant does Albertine just reach theglasswithout knocking it over? Determine the answer "experimentally" byplaying with the applet.Express your answer in newtons permeter.=95.0 Correct!

PART B

Assuming that Albertine's mass is 60.0 , what is $mu_k$ , thecoefficient of kinetic frictionbetween the chair and the waxed floor? Use = 9.80 for the magnitude of theacceleration due to gravity. Assume that the value of found in Part A has three significantfigures.

Note that if you did not assume that has three significantfigures, it would beimpossible to get three significant figures for $mu_k$ , since the length scale along the bottomof the applet does not allow you to measure distances to thataccuracy with different values of.

I attempted Part B as follows:

After the spring becomes fully unstretched, all the elasticpotential energy will have been converted to kinetic energy.Therefore, the velocity of the blockwill be given by the expressonmv^2 = ke

From this, v comes to be 2.8m/s.
After is has just reached the glass, the displacement will be19.8-5 = 14.8m. Also, the final velocity will be 0. Therefore,using the equations of motion, weget 2.8^2 = 0^2 + 2a(14.8). Fromthis, a come to be 0.265m/s^2.

Now, by applying Newton's law of motion. the resultant force whichin this case is just the frictional force is equal to u(9.8x60) =588u.

Therefore, 588u = ma = 60(0.265)
Therefore, the value of u is 0.0270 (3 significant figures).

I have a feeling that this value is way too low! Can someone pleaseadvice me?

Thank-you very very very much!!

Answer 1

So when Albertine compressed and stretched the spring = potential energy = 1/2 k x^2 When he stop at the end of the course (19,8 m from the spring), he hasno mechanical energy. So just consider the work done by nonconservative force that act upon Albertine in straight motion , which is friction. Friction : Fk= uk . N N = Normal Force = mass . gravity Work = Fk . s = uk . mass . gravity . distance If we put both equation together (potential and work done bynonconservative force) : Potential energy = Work done by nonconservative force 1/2 k x^2 = uk . mass . gravity . d 1/2 . 95 . 5^2 = uk . 60 . 9.8 . 19.8 uk= 0.102 Hope this helps !