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podryga [215]
3 years ago
12

Please can someone solve this physics question with a good explenation.

Physics
1 answer:
zimovet [89]3 years ago
3 0

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

Total time taken is, t_{total}=17.5\ s

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

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

Now, from the graph, the vertical height of the triangles is, v=3.41\ m/s

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

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

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}

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

Therefore, the coefficient of dynamic friction is 0.025.

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