Read the Question Below Please :)

Question

A smooth circular hoop with a radius of 0.700 m is placed flat on the floor. A 0.375-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 10.00 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.

(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.

_______J

(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.

_______rev

Answer 1

Find the initial kinetic energy (KE) and subtract the KE after one revolution to get the energy lost to friction.

1/2 x 0.375 kg x (10 m/s)² = 18.75 j (initial KE)

1/2 x .375 kg x (5.5 m/s) = 5.67 joules (final KE)

The energy lost to friction is 18.75 - 5.67 = 13.08 joules per one revolution.

To find how many revolutions it makes before stopping, use the following calculation:

13.08/ 1 rev. x #rev = initial KE Solve for #rev; this gives you the #revolutions for the initial energy to be used up.

#rev. = 18.75 j / 1 rev /13.08 j = 1.43 rev.