Answer:
W= 4.4 J
Explanation
Elastic potential energy theory
If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:
F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.
As the force is variable to calculate the work we define an average force
[tex]F_{a} =\frac{F_{f}+F_{i} }{2}[/tex] Formula (2)
Ff: final force
Fi: initial force
The work done on the spring is :
W = Fa*Δx
Fa : average force
Δx : displacement
[tex]W = F_{a} (x_{f} -x_{i} )[/tex] :Formula (3)
[tex]x_{f}[/tex] : final deformation
[tex]x_{i}[/tex] :initial deformation
Problem development
We calculate Ff and Fi , applying formula (1) :
[tex]F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N[/tex]
[tex]F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N[/tex]
We calculate average force applying formula (2):
[tex]F_{a} =\frac{15.4N+6.2N}{2} = 11 N[/tex]
We calculate the work done on the spring applying formula (3) : :
W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J
Work done in stages
Work is the change of elastic potential energy (ΔEp)
W=ΔEp
ΔEp= Epf-Epi
Epf= final potential energy
Epi=initial potential energy
[tex]E_{pf} =\frac{1}{2} *k*x_{f}^{2}[/tex]
[tex]E_{pi} =\frac{1}{2} *k*x_{i}^{2}[/tex]
[tex]E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J[/tex]
[tex]E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J[/tex]
W=ΔEp= 5.39 J-0.99 J = 4.4J
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