Question
Please show work. Thanks!
A small 0.200 kg package is released from rest at point A on a track that is one quarter of a circle with radius 1.60m as shown. The package slides down the track and reaches pioint B with a speed of 4.80 m/s. From point B, it slideson a level surface a distance of 3.00 m to point C, where is comes to rest.
a. What is the kinetic energy and potential energy at point A?
b. What is the kinetic energy and potential energy at point B?
c. What is the kinetic energy and potential energy at point C?
d. What is the coeffiecient of kinetic friction along the flat surface?
e. How much work is done on the package by friction as it slides down circular arc from A to B?
Explanation / Answer
Given: mass of the particle m = 0.200 kg radius of circular path r = 1.60 m a. At point A the particle is in rest postion so kinetic energy is zero. potential energy P.E. = mgh = (0.200 kg )(9.8 m/sec2)(1.60 m) = 3.136 J b. velocity at point B. v = 4.80 m/sec kinetic energy at this point K.E. = 1/2 mv2 =1/2 * 0.2 kg * (4.8 m/sec)2 = 2.304 J due to conservation energy P.E = 3.136 - 2.304 J = 0.832 J C. At point C the body is at rest postion the energy is zero. D. acceleration of the block from B to C v 2-u2 = 2as final velocity is zero. a = u2/2s = (4.8 m/sec)2/(2*3) = 3.84 m/sec2 k = a/g = (3.84 m/sec2) /(9.8m/sec2 ) = 0.391 E. work done change in kinetic energy W = 2.304 J - 0 = 2.304 J = 2.304 J due to conservation energy P.E = 3.136 - 2.304 J = 0.832 J C. At point C the body is at rest postion the energy is zero. D. acceleration of the block from B to C v 2-u2 = 2as final velocity is zero. a = u2/2s = (4.8 m/sec)2/(2*3) = 3.84 m/sec2 k = a/g = (3.84 m/sec2) /(9.8m/sec2 ) = 0.391 E. work done change in kinetic energy W = 2.304 J - 0 = 2.304 J