The figure shows a 0.41-kg block sliding from A to B along a frictionless surfac
ID: 1624289 • Letter: T
Question
The figure shows a 0.41-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 12 J, and the heights of A and B are 12.0 and 7.0 m above the ground, respectively. a. How fast is the block moving when it reaches B? b. Find the coefficient of kinetic friction between the horizontal surface BC and the block if the distance between points B and C is 21.0 m.Explanation / Answer
A
Potential energy of block at A, (P.E.)A = mghA = (0.41)(9.8)(12) = 48.216 J
Kinetic energy of block at A, (K.E.)A = 0.5*m*vA2 = 12 J
Total energy of block at A, EA = 48.216 J + 12 J = 60.216 J
Potential energy of block at B, (P.E.)B = mghB = (0.41)(9.8)(7.0) = 28.126 J
Kinetic energy of block at B, (K.E.)B = 0.5*m*vB2 = 0.205*vB2 = ?
Total energy of block at B, EB = 28.126 + 0.205*vB2
Using energy conservation,
28.126 + 0.205*vB2 = 60.216
vB = 12.511 m/s
B
As kinetic energy at B, (K.E.)B = 0.5*m*vB2 = 0.205*vB2 = 32.09 J
Kinetic energy at C, (K.E.)C = 0 J
So change in kinetic energy = 32.09 J
Also amount of work done by block in going from B to C, W = f*BC = mg*21 = (0.41)(9.8)21 = 84.378*
Using work energy theorem
84.378* = 32.09
= 0.38