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

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

in progress 0
Ngọc Khuê 3 years 2021-08-04T20:19:50+00:00 1 Answers 9 views 0

Answers ( )

  1. taiduc
    0
    2021-08-04T20:21:02+00:00
    Reply

    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    

                                                   

                                     

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )