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Elenna [48]
4 years ago
11

Explain why the roller coaster’s potential energy is greater at point 1 than at point 4.

Physics
2 answers:
Virty [35]4 years ago
3 0

Potential energy is directly related to the height of an object. Point 1 is higher than point 4, so the potential energy is greater at point 1.

spayn [35]4 years ago
3 0

The most probable answer would be Point 1 is higher than Point 4.


Potential gravitational energy is computed by the formula:


PEgrav =mgh

m = mass

g = acceleration due to gravity (this is a constant value of 9.8m/s²)

h = height


The roller coaster of course, no matter where in the track it is would have the same mass (Unless someone falls off -_-) Pull of gravity is the same as well.


So the only factor that would affect it is the height. So if the height goes up, the potential energy goes up too. If the height goes down, then the potential energy goes down too.


Hopefully you'll come up with the right answer with this. v(0-0)v


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Approximately 13\; {\rm N \cdot m^{-1}} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

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Let F_{\text{s}} denote the force that this spring exerts on the object. Let x denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant k of this spring would ensure that F_\text{s} = -k\, x.

Note that the mass of the object attached to this spring is m = 540\; {\rm g} = 0.540\; {\rm kg}. Thus, the weight of this object would be m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}.

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, F_{\text{s}} = 5.230\; {\rm N}.

The spring in this question was stretched downward from its equilibrium by:

\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}.

(Note that x is negative since this displacement points downwards.)

Rearrange Hooke's Law to find k in terms of F_{\text{s}} and x:

\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}.

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