Answer:
Explanation:
The formula for the work done on a spring is:
[tex]W = \frac{1}{2}kx^{2}[/tex]
where k is the spring constant and x is the change in length of the string
For the first statement,
x = (a-1.1), W = 5
=> [tex]5 = \frac{1}{2}k(a-1.1)^{2}[/tex]
Now making k the subject of formula, we have:
[tex]k = \frac{10}{(a-1.1)^2} ---------------- (A)[/tex]
For the second statement,
x = (a-4.8), W = 9
=> [tex]9 = \frac{1}{2}k(a-4.8)^{2}[/tex]
Now making k the subject of formula, we have:
[tex]k = \frac{18}{(a-4.8)^{2} } ------------------------- (B)[/tex]
Equating A and B since k is constant, we have:
[tex]\frac{10}{(a-1.1)^{2} } = \frac{18}{(a-4.8)^{2} }[/tex]
solving for the value of a
[tex]8a^{2} +56.4a-208.62=0[/tex]
solving for a, we get:
a = 2.6801 or -9.7301
but since length cannot be negative, a = 2.68m
substituting the value of a in equation B, we have:
[tex]k=\frac{18}{(2.68-4.8)^{2} }[/tex]
k = 4.005
