A block slides from rest, along a track with an elevated left end, a flat centra
ID: 1322483 • Letter: A
Question
A block slides from rest, along a track with an elevated left end, a flat central part, into a relaxed spring, as shown in the figure. The curved portion of the track is frictionless, as well as the first portion of the flat part of L = 10 cm. The coefficient of kinetic friction between the block and the only rough part, D = 10 cm, is given by ?k = 0.20. Let the initial height of the block be h = 40 cm, its mass be m = 2.5 kg, and the spring constant k = 320 N/m.
While the block slides through the flat central part of length L find:
The work done on the block by the gravitational force, the work done on the block by the normal force, the work done on the block by the frictional force, and the speed of the puck right before it reaches the beginning of the rough central part of length D.
A block slides from rest, along a track with an elevated left end, a flat central part, into a relaxed spring, as shown in the figure. The curved portion of the track is frictionless, as well as the first portion of the flat part of L = 10 cm. The coefficient of kinetic friction between the block and the only rough part, D = 10 cm, is given by ?k = 0.20. Let the initial height of the block be h = 40 cm, its mass be m = 2.5 kg, and the spring constant k = 320 N/m. While the block slides through the flat central part of length L find: The work done on the block by the gravitational force, the work done on the block by the normal force, the work done on the block by the frictional force, and the speed of the puck right before it reaches the beginning of the rough central part of length D.Explanation / Answer
a)
The work done on the block by the gravitational force
W=mgh
=2.5*9.8*40*10-2
=9.8J
b)
. The work done on the block by the normal force.
=0
because normal force is perpendicular to displacement
c)
The work done on the block by the frictional force
The curved portion of the track is frictionless, as well as the first portion of the flat part of L = 10 cm
w=0 where uk=0
d)
The speed of the block right before it reaches the beginning of the rough central part of length D.
potential energy=kinetic energy
mgh=0.5(mvf2-mvi2)
vf=sqrt(2gh)
=sqrt(2*9.8*40*10-2)
=sqrt(7.84)
=2.8m/s
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