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, [tex]D=29.8\ m[/tex]
Total time taken is, [tex]t_{total}=17.5\ s[/tex]
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:
[tex]v_{avg}=\frac{D}{t_{total}}=\frac{29.8}{17.5}=1.703\ m/s[/tex]
Now, we know that, average speed can also be expressed as:
[tex]v_{avg}=\frac{v_i+v_f}{2}\\1.703=\frac{v+0}{2}\\v=2\times 1.703=3.41\ m/s[/tex]
Now, from the graph, the vertical height of the triangles is, [tex]v=3.41\ m/s[/tex]
The deceleration is given as the slope of the line from time 3.5 s to 17.5 s.
Therefore, deceleration is:
[tex]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[/tex]
Frictional force is the net force acting on the stone. Frictional force is given as:
[tex]f=\mu_dN\\Where, \mu_d\rightarrow \textrm{coefficient of dynamic friction}\\N\rightarrow \textrm{Normal force}\\N=mg\\\therefore f=\mu_dmg[/tex]
Now, from Newton's second law, net force is equal to the product of mass and acceleration.
Therefore,
[tex]\mu_dmg=ma\\\mu_d=\frac{a}{g}[/tex]
Plug in 0.244 for 'a' and 9.8 for 'g'. This gives,
[tex]\mu_d=\frac{a}{g}=\frac{0.244}{9.8}=0.025[/tex]
Therefore, the coefficient of dynamic friction is 0.025.
