The figure shows a 0.412-kg block sliding from A to B along a frictionless surfa
ID: 1437923 • Letter: T
Question
The figure shows a 0.412-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 44.0 J, and the heights of A and B are 11.0 and 7.10 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, you need to note the change in the energy form. As the block slides down from A to B, the extra potential energy it would have at A because of the height gets transformed to kinetic energy by the time it reaches B. Once it starts towards C, the friction would dissipate all of its energy over the distance it travels.
We will use the above to form the energy equation and then find the values required.
A.) At A, the block would have both kinetic energy, 44J, and potential energy, while at B, all of the Potential gets tranformed to KE [We choose the level at which BC is present to be the reference height for PE calculations]
That is, 44 + mgH = KE(at B)
KE(at B) = 44 + 0.412 x 9.81 x 3.9 = 59.763 J
Part B.) Now all the KE the block has at point B, is dissipated by the kinetic friction acting on it while it travels over BC.
Hence the work done by the kinetic frictional force would be equal to the KE at B.
Therefore, the work done by friction = 59.763 J