Suppose a 118kg watermelon is held 5m above the ground before sliding along a frictionless ramp to the ground. How high above the ground is the watermelon at the moment it's kinetic energy is 4,610J
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This question involves the concepts of kinetic energy, potential energy, and the law of conservation of energy.
The watermelon is "1.02 m" above the ground.
The total energy of the watermelon can be found by its potential energy at the highest point:
Total Energy = mgh
where,
Therefore,
Total Energy = (118 kg)(9.81 m/s²)(5 m)
Total Energy = 5787.9 J
Now, according to the law of conservation of energy, at the given point:
Total Energy = Kinetic Energy + Potential Energy
5787.9 J = 4610 J + mgh'
[tex]h'=\frac{5787.9\ J-4610\ J}{(118\ kg)(9.81\ m/s^2)}[/tex]
h' = 1.02 m
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