Isthere an outward force in circular motion
2 answers:
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A) 5000 N/m
The force constant of the spring can be found by using the expression for the elastic potential energy stored in the spring (which is equal to the work done on it):
where
k is the spring constant
x is the stretching/compression of the spring
In this problem, we have
W = 16.0 J is the work done
x = 8.00 cm = 0.08 m is the stretching
Substituting into the formula and re-arranging it, we find
B) 400 N
The magnitude of the force needed to stretch the spring by x = 8.00 cm = 0.08 m is given by Hook's law:
where k=5000 N/m as we found previously. Substituting x=0.08 m, we find:
C) 4 J
The work done to compress the spring by x=4.00 cm=0.04 m is given by the same formula used for part A:
where in this case, k=5000 N/m and x=0.04 m. Substituting, we find
D) 200 N
As we did in part B), the force needed to stretch this distance is given by Hook's law:
where in this case, k=5000 N/m and x=0.04 m. Substituting, we find
Answer:
h = v₀² / 2g , h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²