Answer:
k = 513.71 N/m
Explanation:
To solve this problem we need the concept of Kinetic Energy and Work. Let's start
We know that the Kinetic Energy of the Block is given by
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}1.8*(2.5)^2[/tex]
[tex]KE = 5.625J[/tex]
We can now find the Work made for the Frictional Force, that is
[tex]W = F_r d[/tex]
[tex]W = \mu_k mgd[/tex]
[tex]W = 0.56 * (1.8)(9.8)(0.13)[/tex]
[tex]W = 1.2841J[/tex]
For Energy conservation equation we have
[tex]U = KE-W[/tex]
[tex]U = 5.625J-1.2841[/tex]
[tex]U = 4.3409[/tex]
We can now find k, through the Enery equation which says,
[tex]U = \frac{1}{2}kd^2[/tex]
[tex]4.3409 = \frac{1}{2} k *(0.13)^2[/tex]
Re-arrange for k
[tex]k = \frac{4.3409}{8.45*10^{-3}}[/tex]
[tex]k = 513.71 N/m[/tex]
