FULLSCREEN PRINTER VERSaON BACK NEXT Chapter 06, Problem 90 The figure shows a 0
ID: 1437727 • Letter: F
Question
FULLSCREEN PRINTER VERSaON BACK NEXT Chapter 06, Problem 90 The figure shows a 0.379-kg block sliding from A to B along a frictionless surface. When the block reaches B, it continues to slide along the horizontal surface BC where the kinetic frictional force acts. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A is 43.0 J, and the heights of A and Bare 11.4 and 8.30 m above the ground, respectively. (a) What is the value of the kinetic energy of the block when it reaches B? (b) How much work does the kinetic frictional force do during the BC segment of the trip?Explanation / Answer
Using conservation of energy between A and B
KEA + PEA = KEB + PEB
43 + mg (11.4) = KEB + mg (8.30)
43 + (0.379) (9.8) (11.4) = KEB + (0.379) (9.8) (8.30)
KEB = 54.5 J
work done by kinetic frictional force = final kinetic energy - kinetic energy at B = 0 - 54.5 = 54.5 J
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