A bungee jumper of mass 60 kg jumps from a bridge tied to an elastic rope which becomes taut after he falls 10 m. Consider the jumper when he has fallen another 10 m and is travelling at 15 m/s.
Work out how much energy is stored in the rope. Take g=10 m/s^2 and ignore air resistance.
I need an answer, but I also need to know how to solve it. Thanks in advance.
6 answers
energy stored: mgh=60g*20
It's from a textbook, the answer I'm supposed to get is 5250 Joules..
60*g*20 = 12,000 J or gravitational energy will have been converted to spring and kinetic energy at that point. The kinetic part is (1/2)MV^2 = 6750 J (if you insist on using g = 10 m/s^2).
The remaining energy that must be stored in the spring is 12,000 - 6750 J.
That will give you your book answer.
The remaining energy that must be stored in the spring is 12,000 - 6750 J.
That will give you your book answer.
The total energy= mgh+ 1/2mv^2
= (60*10*10)+ (60*225)/2
= 6000 + 6750
= 12750 J
= (60*10*10)+ (60*225)/2
= 6000 + 6750
= 12750 J
The h should be measure from the centre of gravity. so h is half than the total distance.
Firstly you find the strain energy in the spring:
Mgh=strain energy
M: 60 kg
G: 10 m/s^2
H: 20 m
Strain energy: 60*10*20= 12000
Because you have to ignore air resistance, you just subtract the kinetic energy from the strain energy to find the energy stored in the rope
K.E = 0.5mv^2
M= 60 kg
V= 15 m/s
K.E= 0.5*60*15^2= 6750
Therefore energy in rope = 12000-6750= 5250J
Mgh=strain energy
M: 60 kg
G: 10 m/s^2
H: 20 m
Strain energy: 60*10*20= 12000
Because you have to ignore air resistance, you just subtract the kinetic energy from the strain energy to find the energy stored in the rope
K.E = 0.5mv^2
M= 60 kg
V= 15 m/s
K.E= 0.5*60*15^2= 6750
Therefore energy in rope = 12000-6750= 5250J
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