At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
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Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
(A) Velocity will be 1.88 m/sec
(b) Force will be 187.45 N
Explanation:
We have given work done = 4780 j
Distance d = 25.5 m
(A) Mass of the truck m = ![m=2.70\times 10^3kg](https://tex.z-dn.net/?f=m%3D2.70%5Ctimes%2010%5E3kg)
We know that kinetic energy is given by
![KE=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=KE%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
So ![v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2KE%7D%7Bm%7D%7D%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes%204780%7D%7B2.7%5Ctimes%2010%5E3%7D%7D%3D1.88m%2Fsec)
(B) We know that work done is given by
W = Fd
So ![F=\frac{W}{d}=\frac{4780}{25.5}=187.45N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BW%7D%7Bd%7D%3D%5Cfrac%7B4780%7D%7B25.5%7D%3D187.45N)
Answer:
h = 0.46 m
Explanation:
According to the law of conservation of energy:
Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster
![mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}](https://tex.z-dn.net/?f=mgh%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5C%5C2gh%20%3D%20v%5E2%5C%5C%5C%5Ch%20%3D%20%5Cfrac%7Bv%5E2%7D%7B2g%7D)
where,
h = height = ?
v = speed at bottom = 3 m/s
g = acceleration due to gravity = 9.81 m/s²
Therefore,
![h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B%283%5C%20m%2Fs%29%5E2%7D%7B%282%29%289.81%5C%20m%2Fs%5E2%29%7D)
<u>h = 0.46 m</u>