A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo
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For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J
2) The potential energy at <u>point A</u> is 19620 J
3) The kinetic energy at <u>point B</u> is 10010 J
4) The potential energy at <u>point C</u> is zero
5) The kinetic energy at <u>point C</u> is 19820 J
6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s
1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The<em> total energy</em> is:
Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.
2) The <em>potential energy</em> at point A is:
Then, the potential energy at <u>point A</u> is 19620 J.
3) The <em>kinetic energy</em> at point B is the following:
Since
we have:
Hence, the kinetic energy at <u>point B</u> is 10010 J.
4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.
5) The <em>kinetic energy</em> of the roller coaster at point C is:
Therefore, the kinetic energy at <u>point C</u> is 19820 J.
6) The <em>velocity</em> of the roller coaster at point C is given by:
Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.
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