Applying Energy Conservation With Nonconservative Forces

Consider the situation shown in the figure below, where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N.

The baseball player slides to a stop in a distance \(d\). In the process, friction removes the player’s kinetic energy by doing an amount of work \(\text{fd}\) equal to the initial kinetic energy.

Strategy

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because \(\mathbf{\text{f}}\) is in the opposite direction of the motion (that is, \(\theta =\text{180º}\), and so \(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =-1\)). Thus \({W}_{\text{nc}}=-\text{fd}\). The equation simplifies to

\(\cfrac{1}{2}{{\text{mv}}_{i}}^{2}-\text{fd}=0\)

or

\(\text{fd}=\cfrac{1}{2}{{\text{mv}}_{i}}^{2}\text{.}\)

This equation can now be solved for the distance \(d\).

Solution

Solving the previous equation for \(d\) and substituting known values yields

\(\begin{array}{lll}d& =& \cfrac{{{\text{mv}}_{i}}^{2}}{2f}\\ & =& \cfrac{(\text{65.0 kg})(6\text{.}\text{00 m/s}{)}^{2}}{(2)(\text{450 N})}\\ & =& \text{2.60 m.}\end{array}\)

Discussion

The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.

Suppose that the player from the example above is running up a hill having a \(5\text{.}\text{00º}\) incline upward with a surface similar to that in the baseball stadium. The player slides with the same initial speed, and the frictional force is still 450 N. Determine how far he slides.

The same baseball player slides to a stop on a \(5.00º\) slope.

Strategy

In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at zero height, to the final mechanical energy he has by moving through distance \(d\) to reach height \(h\) along the hill, with \(h=d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}5.00º\). This is expressed by the equation

\(\text{KE}{}_{\text{i}}\text{}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}\text{.}\)

Solution

The work done by friction is again \({W}_{\text{nc}}=-\text{fd}\); initially the potential energy is \({\text{PE}}_{i}=\text{mg}\cdot 0=0\) and the kinetic energy is \({\text{KE}}_{i}=\cfrac{1}{2}{{\text{mv}}_{i}}^{2}\); the final energy contributions are \({\text{KE}}_{f}=0\) for the kinetic energy and \({\text{PE}}_{f}=\text{mgh}=\text{mgd}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \) for the potential energy.

Substituting these values gives

\(\cfrac{1}{2}{{\text{mv}}_{i}}^{2}+0+(-\text{fd})=0+\text{mgd}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta .}\)

Solve this for \(d\) to obtain

\(\begin{array}{lll}d& =& \cfrac{(\cfrac{1}{2}){{\text{mv}}_{\text{i}}}^{2}}{f+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }\\ & =& \cfrac{\text{(0.5)}(\text{65.0 kg})(\text{6.00 m/s}{)}^{2}}{\text{450 N}+(\text{65.0 kg})({\text{9.80 m/s}}^{2})\phantom{\rule{0.25em}{0ex}}{\text{sin (5.00º)}}^{}}\\ & =& \text{2.31 m.}\end{array}\)

Discussion

As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance \(d\) that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy instead, we need only consider the gravitational potential energy \(\text{mgh}\), without combining and resolving force vectors. This simplifies the solution considerably.