# Answer to Question #128092 in Mechanics | Relativity for John

Question #128092
A 50 kg child climbs a 11.5-meter ladder to the top of a slide in 60 seconds.

a. As the boy slides down the slide and reaches the bottom, his speed is 12 m/s and the distance traveled along the slide is 108 m. Determine the average frictional force, in [N], acting on the boy?

b. Find the average power in [W].

c. As the child slides down the slide and reaches the bottom, his speed is 12 m/s and the distance traveled along the slide is 108 m. Find the mechanical energy, in [J], lost by the child in sliding down the slide.

d. Find the gained potential energy in [J].
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2020-08-04T16:00:22-0400

a. Let us calculate the potential energy of the boy when he is on the top of the slide ("g\\approx 9.8\\,\\mathrm{N\/kg}" ).

"E_p = mgh = 5635\\,\\mathrm{J}."

The final kinetic energy is

"E_k = \\dfrac{mv^2}{2} = 3600\\,\\mathrm{J}."

The inequality of energies is due to the frictional force. The difference between the energies is

"\\Delta E = E_p-E_k = 2035\\,\\mathrm{J}."

We calculate the average frictional force as

"F_f = \\dfrac{\\Delta E}{S} = \\dfrac{2035\\,\\mathrm{J}}{108\\,\\mathrm{m}} \\approx 18.8\\,\\mathrm{N}".

b. The average power while climbing is

"W = \\dfrac{E_p}{\\Delta t} = \\dfrac{5635}{60} \\approx 93.9\\,\\mathrm{W}."

c. The lost mechanical energy is "\\Delta E = 2035\\,\\mathrm{J}."

d. The gained potential energy is "E_p = 5635\\,\\mathrm{J}."

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