81 Jumping

Work-Energy Principle

How do we calculate the total work when more than one force acts on an object as it moves, so that each force is doing work? What if the forces point in opposite directions so one does positive work and the other does negative work? In this case we calculate the net work done by each force and add them up (keeping negative works as negative) to get the net work. Alternatively, add up the forces, including directions, to find the size and direction of the net force and then multiply by the distance over which the net force is applied to get the net work.  Either way will give you the same answer, which will be the net work. The net work tells us how much energy is transferred into or out of the kinetic energy, causing a change in kinetic energy (\Delta KE). Everything we have discussed so far can be summed up by the work-energy principle: The change in kinetic energy of a system is equal to the net work on the system, or  written as an equation:

(1)   \begin{equation*} W_{net} = \Delta KE \end{equation*}


(2)   \begin{equation*} W_{net} = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

or in terms of the net force:

(3)   \begin{equation*} F_{net}dcos\theta = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \end{equation*}

Everyday Examples: Jumping

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

First we calculate the work done by their legs.

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

Gravity was acting on them during the launch phase as well, so we need to calculate the work done by gravity, which acts in the opposite direction to motion ($\theta = 180):

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

Adding up the works to get the net work:

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

The work-energy principle tells us to set the change in kinetic energy equal to the net work. 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 isolate the final velocity at the end of the launch phase (as the person leaves the ground)

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

We can see that we need the persons mass. We just need to 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 insert the mass:

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

Finally we take the square root of the result to find the final velocity:

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

Reinforcement Exercises



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