A roller coaster car of mass 1000kg is riding along a track from point A to point C. (point A is 75m high, B is 50m high and C is 100m high, so that there are a total of three hills.) the final velocity at point C is 10m/s

a) Assume no friction what is the speed of the car at the top of hill A?

intial = final
1/2mv_a^2 +mgh_a = 1/2mv_c^2 + mgh_c
1/2V_a^2 + 735 = 1048.02
V_a = 25 m/s

b) what is the max speed the car can achieve along the track? If the speed at point B is 33.2m/s.

What would be the equation for this part of the problem?

c) assume the car starts with the same speed as a) and there is friction present. The car does 50000J of work while the car moves from A to C, will the car still be able to ascend to the top of hill C? At what speed is the car at the top of C?

W= change in KE
and I got V_c = 26.5m/s

Already asked and answered.

I added the speed for point B is that I can get part b). So how would I find the max speed?

b) To find the maximum speed the car can achieve along the track, we need to determine the point where the car will have its maximum kinetic energy. This occurs at the bottom of each hill.

At the bottom of each hill, all the potential energy is converted into kinetic energy:

1/2mv_b^2 + mgh_b = 1/2mv_c^2 + mgh_c

Where:
v_b is the velocity at the bottom of hill B (33.2 m/s)
g is the acceleration due to gravity (9.8 m/s^2)
h_b is the height at the bottom of hill B (50 m)
h_c is the height at the bottom of hill C (100 m)

Plugging in the values:

1/2(1000 kg)v_b^2 + (1000 kg)(9.8 m/s^2)(50 m) = 1/2(1000 kg)v_c^2 + (1000 kg)(9.8 m/s^2)(100 m)

Simplifying:

1/2(1000)v_b^2 + 49000 = 1/2(1000)v_c^2 + 98000

Rearranging the equation to solve for v_c:

1/2(1000)v_c^2 = 1/2(1000)v_b^2 - 49000

v_c^2 = v_b^2 - 98000

Taking the square root of both sides:

v_c = sqrt(v_b^2 - 98000)

c) If the car does 50000 J of work while moving from point A to point C, we can calculate the change in kinetic energy using the work-energy theorem:

W = change in KE

50000 J = 1/2(1000 kg)v_c^2 - 1/2(1000 kg)v_a^2

Plugging in the values:

50000 = 1/2(1000)v_c^2 - 1/2(1000)(25)^2

Simplify:

50000 = 500v_c^2 - 312500

Rearrange the equation:

500v_c^2 = 312500 + 50000

500v_c^2 = 362500

Divide by 500:

v_c^2 = 725

Take the square root of both sides:

v_c = sqrt(725)

v_c ≈ 26.9 m/s

So, with friction present, the car will still be able to ascend to the top of hill C at a speed of approximately 26.9 m/s.

For part b), to find the maximum speed the car can achieve along the track, you can use the conservation of energy principle. At point B, the car has a height of 50m and a velocity of 33.2m/s.

The equation you can use is:

1/2 * m * v_b^2 + m * g * h_b = 1/2 * m * v_max^2 + m * g * h_max

Here, v_b is the velocity at point B, h_b is the height at point B, v_max is the maximum velocity, and h_max is the maximum height the car can reach.

You can rearrange the equation to solve for v_max:

1/2 * v_max^2 = 1/2 * v_b^2 - g * (h_max - h_b)

Substituting the given values:

1/2 * v_max^2 = 1/2 * 33.2^2 - 9.8 * (100 - 50)

Solving for v_max:

v_max^2 = 33.2^2 - 9.8 * 50
v_max = sqrt(33.2^2 - 9.8 * 50)

For part c), if there is friction present and the car does 50000J of work while moving from point A to point C, you can use the work-energy principle to find the velocity at the top of hill C.

The equation you can use is:

Work = change in kinetic energy (KE)

Given that the work done is 50000J, you can set it equal to the change in KE:

50000 = 1/2 * m * v_c^2 - 1/2 * m * v_a^2

Here, v_c is the velocity at the top of hill C, and v_a is the velocity at the top of hill A (which is 25m/s, as calculated in part a)).

Solving for v_c:

1/2 * m * v_c^2 = 50000 + 1/2 * m * v_a^2
v_c^2 = (50000 + 1/2 * m * v_a^2) * (2 / m)
v_c = sqrt((50000 + 1/2 * m * v_a^2) * (2 / m))

Substituting the given values, you can calculate the velocity at the top of hill C.