A massless spring of constant k = 1000 N/m is fixed on the right side of a level

Question

A massless spring of constant k = 1000 N/m is fixed on the right side of a level track. A block of mass m = 1.50 kg is pressed against the spring, compressing it a distance d =20.0 cm, as in the figure below. The block (initially at rest) is then released and travels toward a circular loop-the-loop of radius R = 1.50 m. The entire track and the loop-the-loop are frictionless, except for the section of track between points A and B. The coefficient of kinetic friction between the block and the track along AB is ?k = 0.250 and the length of AB is 3.00 m. a. Determine the speed of the block at point A. b. Calculate the speed of the block at point B. c. Find the speed of the block at point C. What is the minimum speed needed by the block at point C to make it around the top of the loop-the-loop? Will the block make it?

Explanation / Answer

Potential energy stored in spring = (1/2)k d^2= 500 d^2 That will be the kinetic energy the block has departing the spring KE =(1/2)mv^2 = 500 v^2 I assume that A to B is this next stretch between the spring and the bottom of the ferris wheelie. normal force on track = m g friction force = 1.5 m g work done by friction = 1.5 m g (3) = 4.5 m g so kinetic energy at bottom of loop = 500 d^2 - 4.5 m g Now Loss of energy going up loop = m g h = 1.5 m g Now at the top of the loop for zero force on track: m v^2/r = m g So the total kinetic energy at the bottom of the track must be 3.75 m g so 3.75 m g = 44.5 d^2 - .85 m g 44.5 d^2 = 2.90 m g d^2 = (2.90/44.5)(.5)(9.81) d = .565

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