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After work you and your coworkers go see Evil Kenevil Jr. loop-the-loop with his

ID: 2006745 • Letter: A

Question

After work you and your coworkers go see Evil Kenevil Jr. loop-the-loop with his bicycle over a 5.0 m loop where Jr. and his bicycle together have a mass of 75.0 kg.

c. How much work is done by gravity on the dare devil from the top of the loop to the bottom of the loop? W = ?
d. What is the speed of the dare devil at the bottom of the loop if he was traveling the minimum speed needed to stay in contact with the top of the loop the entire time? V = ?
e. What is the magnitude of the normal force on the dare devil at the bottom of the loop? FN = ?

Explanation / Answer

The mass of Jr. and his bicycle, m = 75 kg The radius of the loop, r = 5 m c) The velocity at the bottom of the loop the loop is given by the formula            v1 = sqrt (5 g r ) = sqrt (5 * 9.8 * 5) = 15.6 m/s The velocity at the top of the loop the loop is given by the formula            v2 = sqrt ( g r ) = sqrt (9.8 * 5) = 7 m/s The work done = change in kinetic energy                          = (1/2)mv1^2 - (1/2)mv2^2 = 0.5m(v1^2 - v2^2)                 W = 0.5 * 75(15.6^2 - 7^2) = 7288.5 J = 7.2885 kJ d) The velocity at the bottom of the loop the loop is given by the formula            v1 = sqrt (5 g r ) = sqrt (5 * 9.8 * 5) = 15.6 m/s The velocity at the bottom of the loop the loop is given by the formula            v1 = sqrt (5 g r ) = sqrt (5 * 9.8 * 5) = 15.6 m/s The velocity at the top of the loop the loop is given by the formula            v2 = sqrt ( g r ) = sqrt (9.8 * 5) = 7 m/s The velocity at the top of the loop the loop is given by the formula            v2 = sqrt ( g r ) = sqrt (9.8 * 5) = 7 m/s e) When the man with bicycle is at the bottom of the loop the loop then the forces acts on him are, Upwards : Normal foce = N Down wards: Weight + centrifugal force = mg + mv1^2/r Since the net force along verticle is zero, the upward force is equal to the downward force.      N =mg + mv1^2/r = (70 * 9.8) + (70*15.6*15.6)/(5) = 4093.04 N So the normal force on him along with the bicycle is 4093.04 N