Question

A block of mass 3.5 kg, sliding on a horizontal plane, is released with a velocity of 3.3 m/s. The blocks slides and stops at a distance of 1.6 m beyond the point where it was released. How far would the block have slid if its initial velocity were quadrupled

Answer 1

Answer:

If the initial velocity were quadrupled the block would slide a distance of $\Delta x_{2} =25.6\ m$ .

Explanation:

We are told the mass of the block is $m=3.5\ kg$, the initial velocity in the first case is $v_{i}=3.3\ \frac{m}{s}$ , the final velocity is $v_{f}=0\ \frac{m}{s}$ and the distance at which the block stops in the first case is $\Delta x_{1}=1.6\ m$ .

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:

                         $F_{normal}-F_{weight}=0\ \Longrightarrow\ F_{normal}=F_{weight}=m.g$      

                       $F_{normal}=m.g=3.5\ kg.\ 9.8\frac{m}{s^{2}}\ \Longrightarrow\ F_{normal}=34.3\ N$

In the horizontal direction the only force we have is the force of kinetic friction is $F_{friction}=\mu_{d}.\ F_{normal}$ . We don't know the value of the coefficient of kinetic friction $\mu_{d}$  but this force is the same in both cases.

We use that the change in kinetic energy is equal to the work done by $F_{friction}$ :

                                                          $W=\Delta K$

                                        $-F_{friction}.\ \Delta x=\frac{1}{2}mv_{f} ^{2}-\frac{1}{2}mv_{i} ^{2}$

Because $v_{f}=0\ \frac{m}{s}$  and $F_{friction}$ is the same in both cases we have that:

                                                    $F_{friction}=\frac{1}{2}\frac{m}{\Delta x}\ v_{i} ^{2}$

$\Delta x_{2}$ 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:

                                             $\frac{1}{2}\frac{m}{\Delta x_{1}}\ (v_{i})^{2}=\frac{1}{2}\frac{m}{\Delta x_{2}}\ (4v_{i})^{2}$

If we manipulate the equation we get that:

                                                     $\Delta x_{2}=16\ \Delta x_{1}$      

                                                  $\Delta x_{2}=16\ . \ 1.6\ m$        

                                                    $\Delta x_{2}=25.6\ m$