Answer:
8m * (μg/v)^2
Explanation:
k, the spring constant = ?
(k in terms of μ, m, g, and v.)
Frictional force = μmg
Note: lost KE is converted to work done against the friction + PE of the spring
1/2mv2 = μmgx + 1/2kx^2....equation i
Cancel the 1/2 on both sides
mv^2 = μmgx + kx^2
Lets recall that:
Due to frictional effect, further enegy will be lost when the spring recoils backward
Therefore
1/2kx^2 = μmgx..... equation ii
Let's substitute 1/2kx^2 in equation I for ii
So we can say that:
1/2mv^2 = (μmgx)+ μmgx
1/2mv^2 = 2 (μmgx)
1/4mv^2 = μmgx
Cancel out m on both sides
1/4v^2 = μgx
Make x subject of the formula
x = (1/4v^2) / (μg)...... equation iii
substitute x to equation ii
But first make k in equation ii subject of the formula
1/2kx^2 = μmgx
k = 2μmg/x
Now substitute x
k = 2μmg / ((1/4v^2) / (μg))
k = 2μmg * ((μg) / (1/4v^2))
k = 8m * (μg/v)^2
8m * (μg/v)^2
