Question

A boy in a wheelchair (total mass $\mathbf{47}\mathbf{.}\mathbf{0}\mathbf{}\mathit{g}$) has speed$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ at the crest of a slope$\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}\mathit{m}$ high and$\mathbf{12}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}$ long. At the bottom of the slope his speed is $\mathbf{6}\mathbf{.}\mathbf{20}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. Assume air resistance and rolling resistance can be modeled as a constant friction force of $\mathbf{41}\mathbf{.}\mathbf{00}\mathbf{}\mathit{N}$. Find the work he did in pushing forward on his wheels during the downhill ride.

Short answer

The work he did in pushing forward on his wheels during the downhill ride is

$W_{By-boy}=168J$

Step-by-step solution

Concept and Formulae

Consider the book moves on an incline that is not frictionless, there is a change in both the kinetic energy and the gravitational potential energy of the book Earth system. Consequently,$\Delta E_{mech}=\Delta K+\Delta U_g=-f_{k}d=-\Delta E_{int}$

In general, if a non-conservative force acts within an isolated system

$\sum W_{otherforces}=\Delta K+\Delta U+\Delta E_{int}\sqrt{2}$

where

$\Delta k=$Change in kinetic energy

$\Delta U=$Change in potential energy

$\Delta E_{int}=$Change in internal energy.

Calculating the work

The boy converts some chemical energy in his body into mechanical energy of the boy-chair-Earth system. During this conversion, the energy can be measured as the work his hands do on the wheels. By using the concept from step (1), write the conservation of energy equation for this situation.

$$\begin{gathered} \sum W_{otherforces}=\Delta K+\Delta U+\Delta E_{int} \\ \Delta K+\Delta U=-\Delta E_{int} \\ -\Delta E_{int}=-f_{k}d \end{gathered}$$

Substitute the initial and final energies:

$$\begin{gathered} \left(k_f-k_i\right)+\left(u_f-u_i\right)+\Delta u_{body}=-f_{k}d \\ {k}_i+u_i+W_{hands-on-wheels}-f_{k}d=k_f \end{gathered}$$

Rearranging and renaming, we have

$$\begin{gathered} \frac{1}{2}m{v}_1^2+mg{y}_1+W_{by-boy}-f_{k}d=\frac{1}{2}m{v}_f^2 \\ {W}_{By-boy}=f\frac{1}{2}m\left(v_f^2-v_i^2\right)-mg{y}_1+f_{k}d \end{gathered}$$

Substitute numerical values

$$\begin{gathered} W_{By-boy}=\frac{1}{2}\left(47kg\right)\left[\left(6.20m/s^2\right)-\left(1.40m/s^2\right)\right]-\left(47kg\right)\left(9.8m/s^2\right)\left(2.60m\right)+\left[\left(41.0N\right)\left(12.4m\right)\right] \\ {W}_{By-boy}=168J \end{gathered}$$

Thus, the required work is 168 J.