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You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, after a bad storm some leaves settled on part of the track causing a 9.4-m length of the track to exert a frictional force of 625 n on the car. to safely make it around the loop, the 50-kg car must have a minimum speed of 7.7 m/s at the top of the loop (point b). how fast should the car be moving initially at point a to ensure that it reaches the top of the loop with the minimum required speed?

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Demiurgos
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
[tex]A: mgh_1+\frac{mv_1^2}{2}[/tex]
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
[tex]W_f=F_f\cdot L[/tex]
So, we know that the car is approaching the point B with the following amount of energy:
[tex]mgh_1+\frac{mv_1^2}{2}- F_fL[/tex]
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
[tex]B: mgh_2+\frac{mv_2}{2}[/tex]
As said before this energy must be the same as the energy of a car approaching the loop:
[tex]mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL[/tex]
Now we solve the equation for [tex]v_1[/tex]:
[tex]v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\ v_1^2=39.23\\ v_1=\sqrt{39.23}=6.26\frac{m}{s}[/tex]

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Answer:

speed at initial point A must be v = 6.3 m/s

Explanation:

Here we know that the speed of car at point B must be 7.7 m/s

now by energy conservation we know that total mechanical energy at point B is given as

[tex]E_2 = \frac{1}{2}mv_2^2 + mgh_2[/tex]

[tex]E_2 = \frac{1}{2}(50)(7.7)^2 + (50)(9.8)(12)[/tex]

[tex]E_2 = 7362.25 J[/tex]

Now let say the initial speed at point A is v so initial total energy is given as

[tex]E_1 = \frac{1}{2}m_1v_1^2 + mgh_1[/tex]

[tex]E_1 = \frac{1}{2}(50)v_1^2 + (50)9.8(25)[/tex]

[tex]E_1 = 25v_1^2 + 12250[/tex]

now we know that here energy is lost due to friction which is given by work done by the friction

[tex]W_f = f.d[/tex]

[tex]W_f = (625)(9.4) = 5875J[/tex]

now we know that

energy loss = E1 - E2

[tex]5875 = (25 v_1^2 + 12250) - 7362.25[/tex]

by solving above equation we have

[tex]v_1 = 6.3 m/s[/tex]

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