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# 7.3: Work-Energy Theorem

Skills to Develop

• Apply the work-energy theorem to find information about the motion of a particle, given the forces acting on it
• Use the work-energy theorem to find information about the forces acting on a particle, given information about its motion

We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion.

Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement: dWnet = $$\vec{F}_{net} \cdotp d \vec{r}$$. Newton’s second law tells us that $$\vec{F}_{net} = m \left(\dfrac{d \vec{v}}{dt}\right)$$, so dWnet = m$$\left(\dfrac{d \vec{v}}{dt}\right) \cdotp d \vec{r}$$. For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is,

$$dW_{net} = m \left(\dfrac{d \vec{v}}{dt}\right) \cdotp d \vec{r} = md \vec{v}\; \cdotp \left(\dfrac{d \vec{r}}{dt}\right) = m \vec{v}\; \cdotp d \vec{v},$$

where we substituted the velocity for the time derivative of the displacement and used the commutative property of the dot product [Equation 2.30]. Since derivatives and integrals of scalars are probably more familiar to you at this point, we express the dot product in terms of Cartesian coordinates before we integrate between any two points A and B on the particle’s trajectory. This gives us the net work done on the particle:

$$\begin{split} W_{net,\; AB} & = \int_{A}^{B} (mv_{x} dv_{x} + mv_{y}dv_{y} + mv_{z}dv_{z} \\ & = \frac{1}{2} m \Big| v_{x}^{2} + v_{y}^{2} + v_{z}^{2} \Big|_{A}^{B} = \Big|\frac{1}{2} mv^{2} \Big|_{A}^{B} = K_{B} - K_{A} \ldotp \end{split} \tag{7.8}$$

In the middle step, we used the fact that the square of the velocity is the sum of the squares of its Cartesian components, and in the last step, we used the definition of the particle’s kinetic energy. This important result is called the work-energy theorem (Figure 7.11).

Work-Energy Theorem

The net work done on a particle equals the change in the particle’s kinetic energy:

##### Significance

We could have used Newton’s second law and kinematics in this example, but the work-energy theorem also supplies an answer to less simple situations. The penetration of a bullet, fired vertically upward into a block of wood, is discussed in one section of Asif Shakur’s recent article [“Bullet-Block Science Video Puzzle.” The Physics Teacher (January 2015) 53(1): 15-16]. If the bullet is fired dead center into the block, it loses all its kinetic energy and penetrates slightly farther than if fired off-center. The reason is that if the bullet hits off-center, it has a little kinetic energy after it stops penetrating, because the block rotates. The work-energy theorem implies that a smaller change in kinetic energy results in a smaller penetration. You will understand more of the physics in this interesting article after you finish reading Angular Momentum.

Learn more about work and energy in this PhET simulation (https://openstaxcollege.org/l/ 21PhETSimRamp) called “the ramp.” Try changing the force pushing the box and the frictional force along the incline. The work and energy plots can be examined to note the total work done and change in kinetic energy of the box.

### Contributors

• Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).