Physics, 19.02.2020 01:36 genyjoannerubiera

A roller coaster of mass 3000.0 kg starts from rest at Point A, which is 33 meters above the bottom of the coaster and proceeds to point C, which is 15 m above the bottom of the coaster. a) What is its speed at C? b) Recall that the formula for critical velocity is: If the radius of the loop is 5.0 m, will the roller coaster have enough kinetic energy at the top of the loop to satisfy the critical velocity criteria?

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Answer from: pr173418

Complete answer

A roller coaster of mass 3000.0 kg starts from rest at Point A, which is 33 meters above the bottom of the coaster and proceeds to point C, which is 15 m above the bottom of the coaster. a) What is its speed at C? b) Recall that the formula for critical velocity is: $v=\sqrt{rg}$ If the radius of the loop is 5.0 m, will the roller coaster have enough kinetic energy at the top of the loop to satisfy the critical velocity criteria?

The roller coaster has enough kinetic energy.

Explanation:

Assuming there are not dissipative forces between the cart and the rails of the roller coaster, we can use conservation of energy between A and C:

E_A=E_B=K_A+U_A=K_C+U_C (1)

with K kinetic energy and U potential gravitational energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with v the velocity and m the mass. If we choose zero of potential energy at the bottom of the roller coaster, the gravitational potential energy is:

U=mgh (3)

with gravitational acceleration and h the height of the cart. Using (2) and (3) on (1):

$\frac{mv_A^2}{2}+mgh_A=\frac{mv_C^2}{2}+mgh_C$

Solving for VB and knowing that velocity on A is zero because the cart starts from rest:

$v_C=\sqrt{(\frac{2}{m})(mgh_A-mgh_C)}=\sqrt{2*9.81(33-15)}=18.78\frac{m}{s}$

b) Critical velocity is $v_{cr}=\sqrt{rg}$ in our case:

$v_{cr}=\sqrt{5*9.81}=7.00 \frac{m}{s}$

So, the critical kinetic energy is:

$K_cr=\frac{(3000.0)(7.0)^2}{2}=73500J$

Now we should check that the value of kinetic energy at 5m is more or equal to that value. We can use the equation used on a)

mgh_A=K_5+mgh_5

solving for K5:

K_5=mg(h_A-h_5)=(3000.0)(9.81)(33-5)=824040 J

It is bigger than Kc, so the roller coaster has enough kinetic energy.

Answer from: Quest

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hope this : )

Answer from: Quest

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

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