The figure shows a 0.398-kg block sliding from A to B along a frictionless surfa
ID: 1483351 • Letter: T
Question
The figure shows a 0.398-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 36.0 J, and the heights of A and B are 11.9 and 5.50 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
Here ,
height of A , hA = 11.9 m
height of B , hB = 5.5 m
kinetic energy at A , KEA = 36 J
b) for the kinetic energy of block at B
KEB = KEA + decrease in potential energy
KEB = 36 + 0.398 * 9.8 * (11.9 - 5.5)
KEB = 61 J
the kinetic energy at B is 61 J
b)
as the block come to rest at C
Using work energy theorum
work done by friction = change in kinetic energy during BC
work done by friction = 0 - 61 J
work done by friction = -61 J
the work done by friction is -61 J