# 7.2 Kinetic energy and the work-energy theorem  (Page 3/7)

 Page 3 / 7

## Determining the work to accelerate a package

Suppose that you push on the 30.0-kg package in [link] with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See [link] .) As expected, the net work is the net force times distance.

Solution for (a)

The net force is the push force minus friction, or ${F}_{\text{net}}\text{= 120 N – 5}\text{.}\text{00 N = 115 N}$ . Thus the net work is

$\begin{array}{lll}{W}_{\text{net}}& =& {F}_{\text{net}}d=\left(\text{115 N}\right)\left(\text{0.800 m}\right)\\ & =& \text{92.0 N}\cdot m=\text{92.0 J.}\end{array}$

Discussion for (a)

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

Strategy and Concept for (b)

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for (b)

The applied force does work.

$\begin{array}{lll}{W}_{\text{app}}& =& {F}_{\text{app}}d\phantom{\rule{0.25em}{0ex}}\text{cos}\left(0º\right)={F}_{\text{app}}d\\ & =& \left(\text{120 N}\right)\left(\text{0.800 m}\right)\\ & =& \text{96.0 J}\end{array}$

The friction force and displacement are in opposite directions, so that $\theta =\text{180º}$ , and the work done by friction is

$\begin{array}{lll}{W}_{\text{fr}}& =& {F}_{\text{fr}}d\phantom{\rule{0.25em}{0ex}}\text{cos}\left(\text{180º}\right)=-{F}_{\text{fr}}d\\ & =& -\left(\text{5.00 N}\right)\left(\text{0.800 m}\right)\\ & =& -\text{4.00 J.}\end{array}$

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

$\begin{array}{lll}{W}_{\text{gr}}& =& 0,\\ {W}_{N}& =& 0,\\ {W}_{\text{app}}& =& \text{96.0 J,}\\ {W}_{\text{fr}}& =& -\text{4.00 J.}\end{array}$

The total work done as the sum of the work done by each force is then seen to be

${W}_{\text{total}}={W}_{\text{gr}}+{W}_{N}+{W}_{\text{app}}+{W}_{\text{fr}}=\text{92}\text{.0 J}.$

Discussion for (b)

The calculated total work ${W}_{\text{total}}$ as the sum of the work by each force agrees, as expected, with the work ${W}_{\text{net}}$ done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.

## Determining speed from work and energy

Find the speed of the package in [link] at the end of the push, using work and energy concepts.

Strategy

Here the work-energy theorem can be used, because we have just calculated the net work, ${W}_{\text{net}}$ , and the initial kinetic energy, $\frac{1}{2}{m{v}_{0}}^{2}$ . These calculations allow us to find the final kinetic energy, $\frac{1}{2}{\text{mv}}^{2}$ , and thus the final speed $v$ .

Solution

The work-energy theorem in equation form is

${W}_{\text{net}}=\frac{1}{2}{\text{mv}}^{2}-\frac{1}{2}{m{v}_{0}}^{2}\text{.}$

Solving for $\frac{1}{2}{\text{mv}}^{2}$ gives

$\frac{1}{2}{\text{mv}}^{\text{2}}={W}_{\text{net}}+\frac{1}{2}{m{v}_{0}}^{2}\text{.}$

Thus,

$\frac{1}{2}{\text{mv}}^{2}=\text{92}\text{.}0 J+3\text{.}\text{75 J}=\text{95.}\text{75 J.}$

Solving for the final speed as requested and entering known values gives

$\begin{array}{lll}v& =& \sqrt{\frac{2\text{(95.75 J)}}{m}}=\sqrt{\frac{\text{191.5 kg}\cdot {m}^{2}{\text{/s}}^{2}}{\text{30.0 kg}}}\\ & =& \text{2.53 m/s.}\end{array}$

Discussion

Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

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