86 Comparing Work-Energy and Energy Conservation*

In the previous chapters we analyzed a person’s jump using the rather than the . Examining that correspondence between these concepts will allow us to learn a few important concepts. Let’s refresh ourselves with that example:

Everyday Examples: Jumping

During a jump a person’s legs might apply a force of 1200 N upward on their while the center of mass moves 0.3 m upward.   Let’s figure out what their launch and hang time will be if the person has a weight of 825 N.

First we calculate the done by their legs.

    \begin{equation*} W_L = F_Ldcos\theta = (1200\,\bold{N})(0.3 \,\bold{m})cos(0^{\circ}) = 360 \,\bold{J} \end{equation*}

was acting on them during the launch phase as well, so we need to calculate the done by gravity:

    \begin{equation*} W_g = F_gdcos\theta = (825\,\bold{N})(0.3 \,\bold{m})cos(180^{\circ}) = -247.5 \,\bold{J} \end{equation*}

The net work is then:

    \begin{equation*} W_{net} = -247.5 \,\bold{J} + 360 \,\bold{J} = 112.5 \,\bold{J} \end{equation*}

The tells us to set the change in equal to the . We will keep in mind that they started at rest, so the initial kinetic energy was zero.

    \begin{equation*} 112.5 \,\bold{J} = \frac{1}{2}mv_f^2 - 0 \end{equation*}

We can see that we need the person’s mass. We just divide their weight by g= 9.8 m/s/s to find it:

    \begin{equation*} m = \frac{825\,\bold{N}}{9.8\,\bold{m/s}} = 84.2 \,\bold{kg} \end{equation*}

We isolate the final speed  at the end of the launch phase (as the person leaves the ground) and insert the mass.

    \begin{equation*} \frac{2 ( 112.5 \,\bold{J})}{84.2 \,\bold{kg}} = v_f^2 \end{equation*}

Then we take the square root the result to find the :

    \begin{equation*} v_f = \sqrt{\frac{2 ( 112.5 \,\bold{J})}{84.2 \,\bold{kg}}} = 1.6 \,\bold{m/s} \end{equation*}

In the first part of the jumping example we calculated the on the object and used the to find the change in . In equation form it looks like this:

    \begin{equation*} W_L +W_g =\Delta KE \end{equation*}

Moving the done by to the other side:

    \begin{equation*} W_L =\Delta KE - W_g \end{equation*}

The looks like this:

    \begin{equation*} W_{on} = \Delta KE +\Delta PE +\Delta TE \end{equation*}

Remembering that we ignored and , and that nothing was permanently deformed, \Delta TE = 0.

    \begin{equation*} W_{on} = \Delta KE +\Delta PE + 0 \end{equation*}

Gravity did negative work because it points down and motion was upward, but the effect of that work was to increase potential energy by transferring some of the person’s into their own. Therefore change in gravitational potential energy should be the negative of the work done by gravity. This work was internal to the Earth-person system, so the only work being done on the person’s from the outside was work done by the legs.  Making these replacements we have:

    \begin{equation*} W_{L} = \Delta KE -W_g \end{equation*}

Which is exactly the we started with. We can either use the on a given object and include work done by resistance to and in the , and say all of that work contributes to changing kinetic energy, or instead use the and instead say that work done by elastic forces and gravity contributes to and instead of . Either way is equivalent, as we have just seen.

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Body Physics: Motion to Metabolism Copyright © by Lawrence Davis is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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