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Answer:
a) the total energy is always the same, b ) k = 509.6 N / m
Explanation:
a) In this system the initial energy is elastic potential energy that when the toy is released is transformed into gravitational potential energy and kinetic energy,
Starting point. Maximum compression
Em₀ = [tex]K_{e}[/tex] = ½ k x²
where x is the compression of the spring
Final point
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g h
as there is no friction the (total) mechanical energy is conserved
Em₀ = Em_{f}
½ k x² = ½ m v² + m g h
When analyzing this expression, the total energy is always the same, what it has is a conservation
* at the half height point it is part kinetic and part gravitational potential
* at the point of maximum height it is totally gravitational potential.
b) Let's use conservation of energy
Em₀ = ½ k x2
[tex]Em_{f}[/tex] = U = (m₀ + m) g h
where m₀ is the mass of the spring and m the applied mass
as there is no friction, energy is conserved
Em₀ = Em_{f}
½ k x² = (m₀ + m) g h
m₀ + m =[tex]\frac{k x^{2} }{2g} \ \frac{1}{h}[/tex]
as the student took several values we can perform the calculations a linear graph of m Vs inverse of h ( 1/h)
m(kg) h(m) 1 / h (m⁻¹)
0.03 0.333 3.0
0.045 0.222 4.5
0.06 0.167 6.0
0.075 0.1333 7.5
0.093 0.111 9.0
See the attachment for the graphic,
m = A h' - mo
where A = [tex]\frac{k x^{2} }{2g}[/tex] is the slope of the graph, h ’= 1/h and m₀ is the mass of the spring
the linear regression values for this equation are
m = 0.0104 h' - 0.0018
we substitute
m₀ = 0.0018 kg
\frac{k x^{2} }{2g} = 0.0104
k = 0.0104 [tex]\frac{2g}{x^{2} }[/tex]
let's calculate
k = 0.0104 2 9.8 / 0.02²
k = 509.6 N / m
