84 Jumping

Work-Energy Principle

How do we calculate the total when more than one acts on an object as it moves, so that each force is doing ? 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 . Alternatively, add up the forces, including directions, to find the size and direction of the 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 is transferred into or out of the , causing a change in kinetic energy (\Delta KE). Everything we have discussed so far can be summed up by the :

(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 even more explicitly:

(3)   \begin{equation*} F_{net, ave}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 while the center of mass moves 0.3 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 = (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 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 :

    \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 isolate the 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 . 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

When the person in the previous example reaches peak height their will be instantaneously zero just as they change direction, so their is also zero at this point.

What was the change in kinetic energy of the person from launch to peak of their jump?

While in the air, the only force acting on them was . The done by gravity must have caused the change in kinetic energy you found above. That work was done by gravity applied over the distance that they rose through the air. Apply the to the in-air section of the jump to determine the peak height of their jump. [Hint: You already know the size of the force of gravity, it’s their weight of 825 N]


Share This Book