Answer:
The coefficient of dynamic friction is 0.025.
Explanation:
Given:
Initial speed after the push is 'v' as seen in the graph.
Final speed of the stone is 0 m/s as it comes to rest.
Total distance traveled is, ![D=29.8\ m](https://tex.z-dn.net/?f=D%3D29.8%5C%20m)
Total time taken is, ![t_{total}=17.5\ s](https://tex.z-dn.net/?f=t_%7Btotal%7D%3D17.5%5C%20s)
Time interval for deceleration is 3.5 to 17.5 s which is for 14 s.
Now, average speed of the stone is given as:
![v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s](https://tex.z-dn.net/?f=v_%7Bavg%7D%3D%5Cfrac%7BD%7D%7Bt_%7Btotal%7D%7D%3D%5Cfrac%7B29.8%7D%7B17.5%7D%3D1.703%5C%20m%2Fs)
Now, we know that, average speed can also be expressed as:
![v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s](https://tex.z-dn.net/?f=v_%7Bavg%7D%3D%5Cfrac%7Bv_i%2Bv_f%7D%7B2%7D%5C%5C1.703%3D%5Cfrac%7Bv%2B0%7D%7B2%7D%5C%5Cv%3D2%5Ctimes%201.703%3D3.41%5C%20m%2Fs)
Now, from the graph, the vertical height of the triangles is, ![v=3.41\ m/s](https://tex.z-dn.net/?f=v%3D3.41%5C%20m%2Fs)
The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.
Therefore, deceleration is:
![a=\frac{\textrm{Vertical height}}{\textrm{Time interval}}\\a=\frac{v-0}{17.5-3.5}\\a=\frac{3.41}{14}=0.244\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5Ctextrm%7BVertical%20height%7D%7D%7B%5Ctextrm%7BTime%20interval%7D%7D%5C%5Ca%3D%5Cfrac%7Bv-0%7D%7B17.5-3.5%7D%5C%5Ca%3D%5Cfrac%7B3.41%7D%7B14%7D%3D0.244%5C%20m%2Fs%5E2)
Frictional force is the net force acting on the stone. Frictional force is given as:
![f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg](https://tex.z-dn.net/?f=f%3D%5Cmu_dN%5C%5CWhere%2C%20%5Cmu_d%5Crightarrow%20%5Ctextrm%7Bcoefficient%20of%20dynamic%20friction%7D%5C%5CN%5Crightarrow%20%5Ctextrm%7BNormal%20force%7D%5C%5CN%3Dmg%5C%5C%5Ctherefore%20f%3D%5Cmu_dmg)
Now, from Newton's second law, net force is equal to the product of mass and acceleration.
Therefore,
![\mu_dmg=ma\\\mu_d=\frac{a}{g}](https://tex.z-dn.net/?f=%5Cmu_dmg%3Dma%5C%5C%5Cmu_d%3D%5Cfrac%7Ba%7D%7Bg%7D)
Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,
![\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025](https://tex.z-dn.net/?f=%5Cmu_d%3D%5Cfrac%7Ba%7D%7Bg%7D%3D%5Cfrac%7B0.244%7D%7B9.8%7D%3D0.025)
Therefore, the coefficient of dynamic friction is 0.025.