how to process a simple loop in WWW::Mechanize to be more efficient?

Determine the minimum compression d of the spring that enables the block to just make it through the loop-the-?

• A massless spring of constant k = 85.2 N/m is fixed on the left side of a level track. A block of mass m = 0.50 kg is pressed against the spring and compresses it a distance of d. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.5 m. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. Given that the coefficient of kinetic friction between the block and the track along AB is µk = 0.31, and that the length of AB is 2.5 m, determine the minimum compression d of the spring that enables the block to just make it through the loop-the-loop at point C. [Hint: The force of the track on the block will be zero if the block barely makes it through the-loop-the-loop.] http://www.webassign.net/sf/p7_57.gif there us the diagram. I don't understand what I am doing wrong; this is the process that I am using: ac=v^2/R 9.8=v^2/R v=3.836 a=(mu)(g) a=3.0411 v(final)^2=v(initial)^2 +2ax 3.0411^2=v(initial)^2 +2(-3.0411)(2.5) v(initial)=4.945077 KE=(.5)(m)(v^2) KE=6.11345 PE=KE PE=(.5)(k)(x^2) 6.11345=(.5)(85.2)(x^2) x=.37882 please help me.

• Answer:

  You will have to account for all the energy involved in this setup. Your mistake here is that you haven't accounted for the loss of energy to friction. Here is the correct way. Since we need the block to complete the loop, we require that at the highest point of the loop, the block remains "just" in contact with the loop. In that case, Weight of the block = Centrifugal force in it mg = mv²/R This will fetch you the velocity required at the highest point. It comes out to be 3.83 m/s If this is the velocity at highest point, then the velocity at the lowest point will be, (v(top))² + 2gR = (v(bottom))² (By PE-KE relations) This gives us v(bottom) = 6.64 m/s Again, if this is the velocity after the rough patch, then the velocity before rough patch can be obtained as KE(initial) + Work done by friction = KE (final) ½Mv² + M(-μg)(AB) = ½M(6.64)² This gives the Kinetic energy initial = 14.82 J This must be the energy in the spring = ½kx² = 14.82 This gives us x = 0.58 m It wasn't easy but I hope you get it