Answer:
If the initial velocity were quadrupled the block would slide a distance of [tex]\Delta x_{2} =25.6\ m[/tex] .
Explanation:
We are told the mass of the block is [tex]m=3.5\ kg[/tex], the initial velocity in the first case is [tex]v_{i}=3.3\ \frac{m}{s}[/tex] , the final velocity is [tex]v_{f}=0\ \frac{m}{s}[/tex] and the distance at which the block stops in the first case is [tex]\Delta x_{1}=1.6\ m[/tex] .
In the second case we are told the initial velocity is quadrupled.
Because the block doesn't move in the vertical direction the sum of forces is:
[tex]F_{normal}-F_{weight}=0\ \Longrightarrow\ F_{normal}=F_{weight}=m.g[/tex]
[tex]F_{normal}=m.g=3.5\ kg.\ 9.8\frac{m}{s^{2}}\ \Longrightarrow\ F_{normal}=34.3\ N[/tex]
In the horizontal direction the only force we have is the force of kinetic friction is [tex]F_{friction}=\mu_{d}.\ F_{normal}[/tex] . We don't know the value of the coefficient of kinetic friction [tex]\mu_{d}[/tex] but this force is the same in both cases.
We use that the change in kinetic energy is equal to the work done by [tex]F_{friction}[/tex] :
[tex]W=\Delta K[/tex]
[tex]-F_{friction}.\ \Delta x=\frac{1}{2}mv_{f} ^{2}-\frac{1}{2}mv_{i} ^{2}[/tex]
Because [tex]v_{f}=0\ \frac{m}{s}[/tex] and [tex]F_{friction}[/tex] is the same in both cases we have that:
[tex]F_{friction}=\frac{1}{2}\frac{m}{\Delta x}\ v_{i} ^{2}[/tex]
[tex]\Delta x_{2}[/tex] is the distance the block will travel until stopping when the initial velocity is quadrupled. If we equal the above equation in the first and second case we have that:
[tex]\frac{1}{2}\frac{m}{\Delta x_{1}}\ (v_{i})^{2}=\frac{1}{2}\frac{m}{\Delta x_{2}}\ (4v_{i})^{2}[/tex]
If we manipulate the equation we get that:
[tex]\Delta x_{2}=16\ \Delta x_{1}[/tex]
[tex]\Delta x_{2}=16\ . \ 1.6\ m[/tex]
[tex]\Delta x_{2}=25.6\ m[/tex]
