In the Figure below, we see a 200 gram block resting on the horizontal portion o
ID: 1591175 • Letter: I
Question
In the Figure below, we see a 200 gram block resting on the horizontal portion of a “track”, whose left end is curved upwards. The horizontal portion of the track is frictionless, except for a 80.0 cm long “rough” portion, shown in the Figure, where the coefficient of kinetic friction is 0.400 . The curved portion of the track is frictionless. The block is held against a spring (which has a stiffness constant of 1,000 N/m), compressing the spring by 10.0 cm. Then the block is released from rest. (NOTE: The block is not attached to the spring; it is just held against the end of the spring.)
A) Find the kinetic energy of the block at the instant when it loses contact with the spring.
Assume that at this instant, the spring is neither compressed nor stretched. (At this instant, the block has not yet reached the 80.0 cm long rough portion of the track where friction is significant.)
B) Find the work done on the block by friction, assuming that the block makes it all of the way through the 80.0 cm long rough portion of the track where friction is significant.
C) Find the work done on the block by the normal force while the block slides along the horizontal portion of the track.
D) Find the work done on the block by gravity while the block slides along the horizontal portion of the track.
E) True or False: While the block slides up along the curved portion of the track,
the work done on the block by the normal force equals zero?
F) How much work is done on the block by gravity as the block slides up to its highest point on the curved portion of the track?
G) Through what vertical distance does the block rise as it slides up to its highest point on the curved portion of the track?
H) Find the change in the gravitational potential energy of the system as the block slides up to its highest point on the curved portion of the track.
Please show all work!
Explanation / Answer
K = 1000 N/m
Compression in String, x = 10.0 cm = 0.1 m
uk = 0.4
m = 200 gm = 0.2 kg
(A)
Potential Energy Stored in the String, = 1/2 Kx^2, This will be equal to K.E of the block at the instant it loses contact.
So,
K.E = 1/2 * KX^2
K.E = 1/2 * 1000 * 0.1^2 J
K.E = 5 J
(B)
Work done on the block by Friction,
W = Force * Distance
W = uk * m*g * 0.8 J
W = 0.4 * 0.2 * 9.8 * 0.8 J
W = 0.6272 J
(C)
Work done by the Normal Force = 0 , as there is no displacement in that direction.
(D)
There is no work done by Gravity during the horizontal motion of the block.