A block slides from rest along a track with an elevated left end and a flat cent
ID: 1322557 • Letter: A
Question
A block slides from rest along a track with an elevated left end and 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 uk = 0.20. Let the initial height of th 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 legnth L find:
(a) The work done on the block by the gravitational force.
(b) The work done on the block by the normal force.
(c) The work done on the block by the frictional force.
(d) The speed of the block 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 and 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 uk = 0.20. Let the initial height of th 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 legnth L find: (a) The work done on the block by the gravitational force. (b) The work done on the block by the normal force. (c) The work done on the block by the frictional force. (d) The speed of the block 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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