Answer:
The block slides 1.5m
Explanation:
To solve this problem we need the concepts relate to work energy, work done, frictional force and normal force.
We know for definition that,
[tex]W = \Delta KE = F*d[/tex]
Where W is the work and \Delta KE the change in Kinetic Energy, F the force and d the distance.
[tex]W = \frac{1}{2}kx^2_f-\frac{1}{2}kx^2_i[/tex]
The expression is relate to the Hook's law where F=kx, being k the elastic constant and x the displacement of the object.
We know as well that Frictional Force is given by
[tex]F_f = \mu N = \mu mg[/tex]
Where [tex]\mu[/tex] is the coefficient of friction, m the mass and g the gravity acceleration.
We can now solve the problem. Using the first equation we have,
[tex]W = \frac{1}{2}kx^2_f-\frac{1}{2}kx^2_i[/tex]
Using the second form for work,
[tex]F_f*d = \frac{1}{2}kx^2_f-\frac{1}{2}kx^2_i[/tex]
Here we know that [tex]x_i[/tex] is zero and F_f is equal to [tex]\mu[/tex] mg
[tex](\mu mg)*d = \frac{1}{2}kx^2_f[/tex]
Re-arrange for d,
[tex]d= \frac{{\frac{1}{2}kx^2_f}}{\mu mg}[/tex]
Replacing the values we have,
[tex]d= \frac{340*18*10^{-2}}{2*(0.25)(1.5)(9.8)}[/tex]
[tex]d = 1.498m \approx 1.5m[/tex]
The block slides 1.5m
