FIGURE 9.5 shows a particle moving in a straight line. A constant force $\vec{F}$, which makes an angle $\theta$ with respect to the particle’s displacement $\text{Δ}\vec{r}$, acts on the particle throughout its motion. We’ve established an s-axis in the direction of motion, and you can see that the force component along the direction of motion is $F_s=F\text{cos}\theta$. According to Equation 9.14, the work done on the particle by this force is

FIGURE 9.5 Work being done by a constant force.

Both F and $\text{cos}\theta$ are constant, so they can be taken outside the integral. Thus

Now $s_{\text{i}}$ and $s_{\text{f}}$ are specific to this coordinate system, but their difference $s_{\text{f}}-s_{\text{i}}$ is $\text{Δ}r$, the magnitude of the particle’s displacement vector. Thus we can write more generally, independent of any specific coordinate system, that the work done by the constant force $\vec{F}$ is

where $\theta$ is the angle between the force and the particle’s displacement $\text{Δ}\vec{r}$.

EXAMPLE 9.2 Pulling a suitcase

A strap inclined upward at a $45\text{°}$ angle pulls a suitcase 100 m through the airport. The tension in the strap is 20 N. How much work does the tension force do on the suitcase?

MODEL Let the system consist of only the suitcase, which we model as a particle.

VISUALIZE FIGURE 9.6 is a before-and-after pictorial representation.

FIGURE 9.6 Pictorial representation of the suitcase.

FIGURE 9.6 Full Alternative Text

SOLVE The motion is along the x-axis, so in this case $\text{Δ}r=\text{Δ}x$. We can use Equation 9.17 to find that the tension does work:

$$W=T\text{(}\text{Δ}x\text{)}\text{cos}\theta = (20\text{N})(100\text{m}\text{)}\text{cos}45\text{°}=1400\text{J}$$

REVIEW Because a person pulls the strap, we would say informally that the person does 1400 J of work on the suitcase.

According to the basic energy model, work can be either positive or negative to indicate energy transfer into or out of the system. The quantities F and $\text{Δ}r$ are always positive, so the sign of W is determined entirely by the angle $\theta$ between the force $\vec{F}$ and the displacement $\text{Δ}\vec{r}$.

TACTICS BOX 9.1 Calculating the work done by a constant force

Full Alternative Text

Workbook Exercises 3–7

EXAMPLE 9.3 Launching a rocket

A 150,000 kg rocket is launched straight up. The rocket motor generates a thrust of $4.0\times {10}^6\text{N}$. What is the rocket’s speed at a height of 500 m?

MODEL Let the system consist of only the rocket, which we model as a particle. Thrust and gravity are constant forces that do work on the rocket. We’ll ignore air resistance and any slight mass loss.

VISUALIZE FIGURE 9.7 shows a before-and-after representation and a free-body diagram.

FIGURE 9.7 Before-and-after representation and free-body diagram of a rocket launch.

FIGURE 9.7 Full Alternative Text

SOLVE We can solve this problem with the energy principle, $\text{Δ}K=W_{\text{tot}}$. Both forces do work on the rocket. The thrust is in the direction of motion, with $\theta =0\text{°}$, and thus

$$W_{\text{thrust}}=F_{\text{thrust}}\left(\text{Δ}r\right)=(4.0\times {10}^6\text{N})(500\text{m})=2.00\times {10}^9\text{J}$$

The gravitational force points downward, opposite the displacement $\text{Δ}\vec{r}$, so $\theta =180\text{°}$. Thus the work done by gravity is

$$\begin{array}{ll}{W}_{\text{grav}}\hfill & =-F_{\text{G}}\left(\text{Δ}r\right)=-mg\left(\text{Δ}r\right)\hfill \\ \hfill & =-(1.5\times {10}^5\text{ kg})(9.8{\text{ m/s}}^2)(500\text{ m})=-0.74\times {10}^9\text{ J}\hfill \end{array}$$

The work done by the thrust is positive. By itself, the thrust would cause the rocket to speed up. The work done by gravity is negative, not because $\vec{F}{}_{\text{G}}$ points down but because $\vec{F}{}_{\text{G}}$ is opposite the displacement. By itself, gravity would cause the rocket to slow down. The energy principle, using $K_{\text{i}}=0$, is

$$\text{Δ}K=\frac{1}{2}m{v}_1{}^2-0=W_{\text{tot}}=W_{\text{thrust}}+W_{\text{grav}}=1.26\times {10}^9\text{ J}$$

Solving for the speed, we find

$$v_1=\sqrt{\frac{2W_{\text{tot}}}{m}}=130\text{ m/s}$$

REVIEW The total work is positive, meaning that energy is transferred to the rocket. In response, the rocket speeds up.