Non-Conservative Forces

As you read, pay attention to Figure 7.15 for a comparison of conservative and non-conservative forces. In Figure 7.15 (a), a rock is being "bounced" on an ideal spring with no friction. The mechanical energy does not change, and the rock will continue bouncing indefinitely. In Figure 7.15 (b), the rock is thrown and lands on the ground. When it hits the ground, its kinetic energy is converted to thermal energy and sound. The rock can not "bounce" back up because its mechanical energy is not conserved.

Gravity is a good example of a conservative force we use a lot in physics. Gravitational force is a conservative force because the work gravity does on an object does not depend on the path the object takes. Consequently, gravity is a good candidate to add into the work-energy theorem, where work is only done by gravity: $W=Fd=mad$

Since the acceleration due to gravity is simply $g$ and the direction of motion due to gravity is in the y-axis, we can further build the equation that represents work due to gravity: $W=mg(\Delta y)=\Delta(mgy)$

Previously, we have discovered that work is also equal to the change in kinetic energy (see Section 7.2). So, we can now combine our equation for work due to gravity and our equation for work with respect to the change in kinetic energy: $\Delta(mgy)=\Delta(\frac{1}{2})mv^{2}$ . The $mgy$ in the equation is called the gravitational potential energy. We define potential energy as stored energy due to a system's position: $PE=mgy$ .

Example 7.10 Calculating Distance Traveled: Sliding Up an Incline

Suppose that the player from Example 7.9 is running up a hill having a $5.00^{\circ}$ 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.

A baseball player slides on an inclined slope represented by a right triangle. The angle of the slope is represented by the angle between the base and the hypotenuse, which is equal to five degrees, and the height h of the perpendicular side of the triangle is equal to d sin 5 degrees. The length of the hypotenuse is d.

Figure 7.18 The same baseball player slides to a stop on a $5.00^{\circ}$ 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 $h=d \sin 5.00^{\circ}$ . This is expressed by the equation

$\mathrm{KE}_{\mathrm{i}}+\mathrm{PE}_{\mathrm{i}}+W_{\mathrm{nc}}=\mathrm{KE}_{\mathrm{f}}+\mathrm{PE}_{\mathrm{f}}$ .

Solution

The work done by friction is again $W_{\mathrm{nc}}=-f d$ ; initially the potential energy is $\mathrm{PE}_{\mathrm{i}}=m g \cdot 0=0$ and the kinetic energy is $\mathrm{KE}_{\mathrm{i}}=\frac{1}{2} m v_{\mathrm{i}}^{2}$ ; the final energy contributions are $\mathrm{KE}_{\mathrm{f}}=0$ for the kinetic energy and $\mathrm{PE}_{\mathrm{f}}=m g h=m g d \sin \theta$ for the potential energy.

Substituting these values gives

$\frac{1}{2} m v_{\mathrm{i}}^{2}+0+(-f d)=0+m g d \sin \theta$ .

Solve this for $d$ to obtain

$\begin{aligned} d &=\frac{\left(\frac{1}{2}\right) m v_{\mathrm{i}}^{2}}{f+m g \sin \theta} \\ &=\frac{(0.5)(65.0 \mathrm{~kg})(6.00 \mathrm{~m} / \mathrm{s})^{2}}{450 \mathrm{~N}+(65.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right) \sin \left(5.00^{\circ}\right)} \\ &=2.31 \mathrm{~m} \end{aligned}$

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 $mgh$ , without combining and resolving force vectors. This simplifies the solution considerably.