A roller coaster powered by a spring launcher follows a frictionless track as shown below. The roller coaster cart has a mass of 500 kg, and the spring launcher has a spring constant of 1500 N/m.
a. If the cart starts at a height h1 = 10m, how much must the spring be compressed for the cart to make it over the hill at h2 if h2=25m?
b. What is the speed of the cart when it reaches h3 if h3=5m (assume that the cart is launched as derived in part a)?
c. What is the total work done by gravity on the cart between h1 and h3?
d. What is the fastest speed the cart reaches anywhere on the track?
If H2 = 25
then the car must have potential energy of
m g H2 = 500(9.81)(25) = 122,625 Joules at start
so
122,625 = m g H1 + (1/2) k x^2
122,625 = 500*9.81*10 + (1/2) k x^2
(1/2) 1500 * x^2 = 73,575
so
x = 9.9 meters
part b later maybe
well at H2 it had total energy of 122,625.
now it goes down 20 meters to 5 meters off ground
so it loses m*9.81*20 joules of potential
so
(1/2) m v^2 = m * 9.81 *20
v= sqrt (2 * 9.81*20)
by the way if something falls from height h it is handy to remember that its speed reaches sqrt(2 g h)
v = 19.8 m/s
H1 = 10
H3 = 5
falls 5 meters in all
m g h = 500 * 9.81 * 5 = 24,525 Joules
The highest speed is at the lowest point.
I have to assume that is H3 of 5 meters
so it started with 122,625 Joules
it then dropped from the 10 meter start to the 5 meter point 3, gaining 24,525 Joules
so
(1/2) m v^2 = 122,625 + 24,525
v^2 = 2(147150) / 500
v = 24.3 m/s
To answer these questions, we'll need to apply the principles of conservation of energy and work-energy theorem.
a. The initial potential energy of the cart at height h1 is given by mgh1, where m is the mass of the cart (500 kg), g is the acceleration due to gravity (9.8 m/s^2), and h1 is the initial height (10 m). The final potential energy at height h2 is given by mgh2, where h2 is the final height (25 m). The change in potential energy is equal to the work done on the cart.
Since the track is frictionless, the only work done on the cart is by the spring launcher. The potential energy stored in a spring is given by (1/2)kx^2, where k is the spring constant (1500 N/m) and x is the compression of the spring. Equating the change in potential energy to the potential energy stored in the spring, we have:
(1/2)kx^2 = mgh2 - mgh1
Substituting the given values, we get:
(1/2)(1500 N/m)x^2 = (500 kg)(9.8 m/s^2)(25 m) - (500 kg)(9.8 m/s^2)(10 m)
Simplifying and solving for x, we can find the amount the spring must be compressed.
b. To find the speed of the cart at height h3, we need to consider the conservation of mechanical energy. At height h2, the total mechanical energy of the system (cart + Earth) is the sum of the kinetic energy and potential energy. At height h3, the total mechanical energy remains the same.
The mechanical energy at height h2 is given by the sum of kinetic and potential energy:
(1/2)mv^2 + mgh2
The mechanical energy at height h3 is given by the sum of the kinetic and potential energy:
(1/2)mv^2 + mgh3
Since the mechanical energy is conserved, we have:
(1/2)mv^2 + mgh2 = (1/2)mv^2 + mgh3
Simplifying and solving for v, we can find the speed of the cart.
c. The work done by gravity on the cart is given by the negative change in potential energy. The change in potential energy is equal to the work done against gravity, which is given by mgh3 - mgh1. Substituting the given values, we can calculate the total work done by gravity.
d. The fastest speed the cart reaches anywhere on the track occurs at the bottom of the hill. At the bottom of the hill, the potential energy is zero, so the total mechanical energy is equal to the kinetic energy: (1/2)mv^2. Using the conservation of mechanical energy, we can find the maximum speed of the cart.
By applying the above principles and calculations, you can find the answers to all the questions.