boy in a wheelchair (total mass 52.0 kg) has speed 1.40 m/s at the crest of a slope 2.35 m high and 12.4 m long. At the bottom of the slope his speed is 6.30 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

According to the law of energy conservation, the work he did in pushing forward on his wheels and the work produced by friction during the downhill ride can be equal to the change of mechanical energy between the top and the bottom of the slope.

Let g = 9.81 m/s2

Let the bottom of the slope be ground 0 for his potential energy, then his potential energy at the top height h = 2.35 m of the slope would be

[tex]E_p = mgh = 52*9.81*2.35 = 1199 J[/tex]

His kinetic energy at the top of the slope is

[tex]E_{kt} = mv_t^2/2 = 52*1.4^2/2 = 50.96 J[/tex]

His kinetic energy at the bottom of the slope would be

[tex]E_{kb} = mv_b^2/2 = 52*6.3^2/2 = 1032 J[/tex]

The work done by friction F = 41N over distance of s = 12.4m is

[tex]W_f = Fs = 41*12.4 = 508.4 J[/tex]

Therefore we can apply the law of energy conservation

[tex]E_p + E_{kt} + W = E_{kb} + W_f[/tex]

[tex]W = E_{kb} + W_f - E_p - E_{kt} = 1032 + 508.4 - 1199 - 50.96 = 290.6 J[/tex]

shahnoorazhar3 shahnoorazhar3 · Answer 2 · 2020-04-01T03:42:41+02:00

Answer:

W_b = 290.6 J

Explanation:

Given:-

- The initial velocity vi = 1.40 m/s

- The final velocity vf = 6.30 m/s

- The Length of the slope, L = 12.4 m

- The height of the slope, h = 2.35 m

- The frictional force, Ff = 41.0 N.

- The mass of the boy , m = 52.0 kg.

Find:-

Find the work he did in pushing forward on his wheels during the downhill ride.

Solution:-

- The work done by the boy on the wheels during his journey down the hill can be modeled by the work-energy principle. Where, the net work done W_net is equal to change in total mechanical energy of the system. We have:

Δ K.E + ΔP.E = W_net

Where,

ΔK.E : Change in kinetic Energy of the system

ΔP.E : Change in potential energy of the system.

- The net work done on the system consists off the work done the boy (W_b) and the work done by the system against resistive forces (W_f).

W_net = W_b - W_f

W_net = W_b - Ff * L

- The change in kinetic and potential energies can be expressed as:

0.5*m*( vf^2 - vi^2) - m*g*h = W_b - Ff * L

0.5*(52.0)*(6.3^2 - 1.40^2) - (52)(9.81)(2.35) = W_b -41 *12.4

-217.802 + 508.4 = W_b

W_b = 290.6 J