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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

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 class=

Respuesta :

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,

  • m = mass = 118 kg
  • g = 9.81 m/s²
  • h = height = 5 m

Therefore,

Total Energy = (118 kg)(9.81 m/s²)(5 m)

Total Energy = 5787.9 J

LAW OF CONSERVATION OF ENERGY

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

Learn more about the law of conservation of energy here:

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