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Ksivusya [100]
3 years ago
7

At which of the following points does a roller coaster have the most potential energy? As it is going down a hill. At the top of

the hill. As it is climbing a hill. At the bottom of the hill.
Physics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

At the top of the hill.

Explanation:

As the roller coaster goes up the hill, kinetic energy (K.E) decreases, gravitational potential energy (G.P.E) increases .

As it reach the top of the hill, K.E becomes zero and G.P.E reaches <em>m</em><em>a</em><em>x</em><em>i</em><em>m</em><em>u</em><em>m</em> .

As it goes down the hill, K.E starts to increase and G.P.E decrease .

At the bottom of the hill, K.E reaches <em>maximum</em> and G.P.E becomes zero .

(Correct me it I am wrong)

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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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