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A 30.0 kg child, initially at rest, slides down a 2.0 m tall slide. The child reaches the bottom of the slide with a speed of 6 m/s. There is friction between the child and the slide. Write a Law of Conservation of Energy equation to represent the transfer of energy from the top of the slide to the bottom. a. b. Use the equation from part (b) to calculate the energy dissipated by friction between the slide and the child? (g = 9.81 m/s) = 13​

Respuesta :

Answer:

Explanation:

Total energy is constant

E = mgh + ½mv² + Fd

At the top of the slide, all energy is potential

E = mgh + 0 + 0

At the bottom of the slide, all potential energy has converted to kinetic and work of friction.

mgh = ½mv² + W

W = mgh - ½mv²

W = 30.0[(9.81)(2.0) - ½6²]

W = 48.6 J

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