The figure shows a 0.415-kg block sliding from A to B along a frictionless surfa
ID: 1370321 • Letter: T
Question
The figure shows a 0.415-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 10.5 and 6.30 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
For A the total energy is the same for B
Ka=44 J (kinetic energy)
Pa (potencial energy) Pa=mgh=(0.415 kg)(9.81 m/s^2)(10.5 m)= 42.74 J
So the total energy for a is E=Ka+Pa= 44+42.74= 86.74 J
the Ea is the same in b so
Eb=86.74 J = Kb+Pb and Pb=mgh=(0.415 kg)(9.81 m/s^2)(6.30)=25.6482 J, so the kinetic in B is
Kb=Eb-Pb=86.74-25.6482=61.0918 J is the kinetic energy of the block in b
B)
The work that the kinetic energy does form B to C is the same that has the block at the begining in B, so
Work=61.0918 J