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Svetach [21]
3 years ago
9

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi

gh hill, then descends 19 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.3 m and that a loaded car will have a maximum mass of 410 kg . For safety reasons, the spring constant should be 12 % larger than the minimum needed for the car to just make it over the top. What is the maximum speed of a 340kg car if the spring is compressed the full amount?
Physics
1 answer:
True [87]3 years ago
4 0

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

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4 years ago
Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2 h, and Car B traveled the dista
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Car A:
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t1 = 2 h
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Car B:
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<span>Two cars traveled equal distances:
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<span>Car B traveled 15 mph faster than Car A:
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</span>v1 * t1 = v2 * t2
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v1 * 2 = (v1 + 15) * 1.5
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8 0
3 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
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And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

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4 0
3 years ago
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Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
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