The results of the experiments should be similar.
Answer:
a)= 29.4J
b)F = 588 N
c)= 60 Kg
Explanation:
Force constant of the spring (k) = 5880 N/m
Change in length of the spring (x) = 25 - 15 = 10 cm 0.1m
This work done on the spring as it is stretched (or compressed) can be recovered. This is stored work that can be used to do work on something else by this spring. That means the stretched (or compressed) spring has energy -- potential energy. This is spring potential energy or elastic potential energy.
a) Work done in pulling the body W = 1/2kx²
= 1/2 (5880)(0.1)2
= 29.4J
b)From Hook's Law,
F = ke
Where F = applied force, k = spring constant, e = extension.
Given: k = 5880 N/m, e = 25-15 = 10 cm = 0.1 m.
Substitute into the formula above
F = 5880(0.1)
F = 588 N
c)By using the formula, F = -kx
Hence mg = kx
Thus m x 9.8 =5880 x0.1
Hence mass of the body
m= 5880 x0.1/9.8
= 60 Kg
Given:
Circumference = 2 m
Angular speed, ω = 1 rev/s = 2π radians/s
If the radius is r, then
2πr = 2
r = 1/π m
The linear (tangential) speed is
v = rω
= (1/π m)*(2π rad/s) = 0.5 m/s
Answer: 0.5 m/s
(a) The final angular velocity of the flywheel after 3 complete revolutions is 4.96 rad/s.
(b) The time taken for the flywheel to make 3 complete revolutions is 5.93 s.
<h3>
Final angular velocity</h3>
The final angular velocity of the flywheel after 3 complete revolutions is determined by applying third kinematic equation as shown below;
θ = 2π (rad/rev) x (3 rev) = 18.85 rad
ωf² = ωi² + 2αθ
ωf² = (1.4)² + 2(0.6)(18.85)
ωf² = 24.58
ωf = √24.58
ωf = 4.96 rad/s
<h3>Time of motion</h3>
The time taken for the flywheel to make 3 complete revolutions is calculated as follows;
ωf = ωi + αt
t = (ωf - ωi)/α
t = (4.96 - 1.4)/0.6
t = 5.93 s
Learn more about time of motion here: brainly.com/question/2364404
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