- #1
AnnieD
- 24
- 0
In Fig.7-11
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_11.gif
we must apply a force of magnitude 82.0 N to hold the block stationary at x=-2.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the block is then again stationary. What are the block's positions, in cm? ((a) positive and (b) negative)
This is what I did, but can anyone confirm if it's correct?
F = -kx
82 = -k(-2)
82 = 2k
41 = k
Convert 8J aka 8N/m to .08N/cm because the final answer has to be in cm.
U = 1/2kx^2
For positive:
82 + .08 = 1/2(41)x^2
solve for x .. 2.00cm
For negative:
82 - .08 = 1/2(41)x^2
solve for x.. 2.00cm
(has to have 3SF)
Something doesn't seem right.. maybe I messed up in converting?
Any help would be much appreciated.. thanks!
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_11.gif
we must apply a force of magnitude 82.0 N to hold the block stationary at x=-2.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the block is then again stationary. What are the block's positions, in cm? ((a) positive and (b) negative)
This is what I did, but can anyone confirm if it's correct?
F = -kx
82 = -k(-2)
82 = 2k
41 = k
Convert 8J aka 8N/m to .08N/cm because the final answer has to be in cm.
U = 1/2kx^2
For positive:
82 + .08 = 1/2(41)x^2
solve for x .. 2.00cm
For negative:
82 - .08 = 1/2(41)x^2
solve for x.. 2.00cm
(has to have 3SF)
Something doesn't seem right.. maybe I messed up in converting?
Any help would be much appreciated.. thanks!