Answer to Question #128092 in Mechanics | Relativity for John

Question

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].

Answer 1

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 $E_p = 5635\,\mathrm{J}.$