A 240 kg roller coaster car Rolls over the crest of the roller coaster with a speed of 2.2m/s.if it falls through a vertical height of 12m down the tracks to the bottom of a trough, how fast will it be travelling?
So here's what I did but I don't think it's right:
Eg=mgh
=240x9.8x12
=28224J
Ek=(1/2)(240)(2.2)
=264J
EM=ek+eg
=28488
So I assumed that h=0 at the bottom of the trough
Eg=9.8x0x240
=0
So therefore ek=28489
28488J=1/2(240)v^2
15.4m/s=v
I'm not sure if this is actually correct?? Thank you so much
“Ek=(1/2)(240)(2.2)
=264J”
This is wrong. Ek = 1/2mv SQUARED. Not 1/2mv.
To solve this problem, you have correctly used the principles of conservation of energy. However, there seems to be a small calculation mistake in your working. Let's go through the steps again:
1. Calculate the potential energy (PE) at the top of the roller coaster crest:
PE = mgh
PE = 240 kg * 9.8 m/s^2 * 12 m
PE = 28,224 J
2. Calculate the kinetic energy (KE) at the top of the roller coaster crest:
KE = (1/2)mv^2
KE = 0.5 * 240 kg * (2.2 m/s)^2
KE = 582.4 J
3. Calculate the total mechanical energy (EM) at the top of the roller coaster crest:
EM = PE + KE
EM = 28,224 J + 582.4 J
EM = 28,806.4 J
At this point, your calculation diverges. Instead of assuming h=0 at the bottom of the trough, we can use the principle of conservation of energy to determine the velocity at the bottom.
4. Calculate the total mechanical energy (EM) at the bottom of the trough:
EM = PE + KE
EM = 0 + KE (since the potential energy is zero at the bottom)
EM = KE
5. Set the total mechanical energy at the bottom equal to the kinetic energy equation:
EM = (1/2)mv^2
28,806.4 J = 0.5 * 240 kg * v^2
6. Solve for v:
v^2 = (2 * 28,806.4 J) / 240 kg
v^2 = 2 * 120.03
v^2 = 240.06
v = sqrt(240.06)
v ≈ 15.49 m/s
So, the roller coaster car will be traveling at approximately 15.49 m/s at the bottom of the trough.