Respuesta :
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
Answer:
a) W = 1815 J
b) W = 135 J
Explanation:
We need to model the force using Hooke's law
We first get the elastic constant, k
A force of 60 N causes an extension of 0.5 m
F = 60 N, e = 0.5 m
F = k * e
60 = 0.5 k
k = 60/0.5
k = 120 N/m
Therefore the force in terms of the extension, x is:
F(x) = 120x
a) Work done in stretching the spring 5.5 m from its equilibrium position
b = 0 m, a = 5.5 m
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex]
[tex]W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5[/tex]
[tex]W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ][/tex]
[tex]W = 1815 J[/tex]
b) Work required to compress the spring 1.5 m from its equilibrium position
[tex]W = \int\limits^a_b {F(x)} \, dx[/tex] b = 0, a = -1.5
[tex]W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]\\[/tex]
W = 135 J
The work done to compress the spring 1.5 m from its equilibrium position is 135 J
