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Tristin_noel
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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 uk = 0.20. Let the initial height of the block be h = 40cm, its mass be m = 2.5 kg, and the spring constant k = 320 N/m. (Picture should be posted below.)
1. While the block slides through the flat central part of length L find:
a) The work done on the block by the gravitational force. = would you use W=mgh ?
b) The work done on the block by the normal force. = 0 bc the normal force is perpendicular to the displacement.
c) The work done on the block by the frictional force. = 0 bc no frictional force present
d) The speed of the block right before it reaches the beginning of the rough central part of length D. =2.8m/s used the eqautopn PE=KE , mgh=1/2m(vf^2-vi^2)
2) The block has now made it to the rough region of length D and will eventually slide head on into the spring at the end of the path. Over this region of length D, calculate:
e) The work done on the block by the gravitational force. = 0 bc perpendicular
f) The work done on the block by the normal force. = 0 bc perpendicular
g) The work done on the block by the frictional force. =-0.49J bc Wfriction=uk(mg)d
h) The speed of the block just as it reaches the spring. =2.73m/s using Vf^2-Vi^2=2(-ukg)d
3) Assuming the block compresses the spring by x, find:
i) The work done by the spring force. would you use W=1/2mVf^2 ?
j) The compression distance, x, of the spring. would you use W=kx and just solve for x?
https://s.yimg.com/hd/answers/i/6ae6d9b8385d4d3eb364c25c31aeb7d5_A.png?a=answers&mr=0&x=1425534324&s=478f148cd3470b7f93d03b19b2553c0b
1. While the block slides through the flat central part of length L find:
a) The work done on the block by the gravitational force. = would you use W=mgh ?
b) The work done on the block by the normal force. = 0 bc the normal force is perpendicular to the displacement.
c) The work done on the block by the frictional force. = 0 bc no frictional force present
d) The speed of the block right before it reaches the beginning of the rough central part of length D. =2.8m/s used the eqautopn PE=KE , mgh=1/2m(vf^2-vi^2)
2) The block has now made it to the rough region of length D and will eventually slide head on into the spring at the end of the path. Over this region of length D, calculate:
e) The work done on the block by the gravitational force. = 0 bc perpendicular
f) The work done on the block by the normal force. = 0 bc perpendicular
g) The work done on the block by the frictional force. =-0.49J bc Wfriction=uk(mg)d
h) The speed of the block just as it reaches the spring. =2.73m/s using Vf^2-Vi^2=2(-ukg)d
3) Assuming the block compresses the spring by x, find:
i) The work done by the spring force. would you use W=1/2mVf^2 ?
j) The compression distance, x, of the spring. would you use W=kx and just solve for x?
https://s.yimg.com/hd/answers/i/6ae6d9b8385d4d3eb364c25c31aeb7d5_A.png?a=answers&mr=0&x=1425534324&s=478f148cd3470b7f93d03b19b2553c0b
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