Work

Work is done when a force is applied, at least partially, in the direction of the displacement of the object. If that force is constant then the work done by the force is the dot product of the force with the displacement:

\[W=\vec{F} \bullet \vec{d}.\]

The net work done on an object (the work of the net force) is equal to the energy added to the object. This is the reason for both the unit of work being the Joule, \(\text{J}\), and the work-kinetic energy theorem:

\[W=\Delta K.\]

Anne spends most of her day napping on a futon. In the reference frame of the Earth, what is the work done to Anne by the futon in one hour? A futon is a couch that can also be folded flat to serve as a bed.

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In the reference frame of the Earth, Anne is sitting still, so her displacement is 0 \(\text{m}\) in any amount of time. Hence the work done in one day is

\[W=\vec{F} \bullet \vec{d} = \vec{F} \bullet \vec{0} = 0 \text{ J}.\]

Newton's laws provide a method for turning force into acceleration, \(F_{net}=ma,\) but this only provides the acceleration at an instant. In order to affect the motion of an object, the force must be applied over some time (impulse) or over a distance (work). For an object with displacement, \(\vec{d}\), only in the x-direction being acted on only by a force in the x direction, the work is simply defined in terms of the magnitudes as

\[W_x=F_x d_x.\]

If there are also some y and z components to the force, the work of each of those forces is

\[W_y=F_y d_y\]

\[W_z=F_z d_z.\]

The net work on an object moving through space is just the sum of these individual components.

\[W_{net} = W_x + W_y + W_z\]

\[W_{net} = F_x d_x + F_y d_y + F_z d_z\]

Since the right side is the definition of the dot product, the left side must be as well.

\[\vec{F}\bullet\vec{d} = F_x d_x + F_y d_y + F_z d_z\]

This dot-product realization also produces another valuable form of the work definition.

\[\vec{F}\bullet\vec{d} = Fd \cos(\theta)\]

A force \(\vec{F} = (3 \text{ N})\hat{x} + (4 \text{ N})\hat{y}\) causes a block to undergo a displacement \(\vec{d} = (2 \text{ m})\hat{y}.\) Find the work done.

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Since the force and displacement are given in terms of their components, use the component-wise definition of work.

\[\vec{F}\bullet\vec{d} = F_x d_x + F_y d_y + F_z d_z = (3 \text{ N})(0 \text{ m}) + (4 \text{ N})(2 \text{ m}) +(0 \text{ N})(0 \text{ m}) = 8 \text{ J}\]

Combining Newton's second law, \(F_{net} = ma,\) with the definition of work in one-dimension gives

\[W_x=F_x d_x = ma_x d_x.\]

The product \(a_x d_x\) shows up in one dimensional kinematics.

\[2 a_x d_x = v_f^2 - v_i^2\]

Divide both sides by 2 to solve for the desired product on the left.

\[a_x d_x = \frac12 v_f^2 - \frac12 v_i^2\]

Finally, multiply both sides by mass to get work on the left.

\[ma_x d_x = \frac12 mv_f^2 - \frac12 mv_i^2\]

Notice, while the left is the definition of work, the right also happens to be the change in kinetic energy.

\[W_{net} = \Delta K\]

Combined with the conservation of energy, \(\Delta K = -\Delta U,\) provides the following relation.

Work of a conservative force \[W_{net} = \Delta K = -\Delta U\]

The kinetic energy of a car decreases from \(2.0 \times 10^6 \text{ J}\) to \(1.0 \times 10^6 \text{ J}\) over a distance of 10 \(\text{m}\) by a braking force. Find the magnitude of that braking force.

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\[W_{net} = \Delta K\] \[Fd = K_f - K_i\] \[F(10\text{ m}) = 1.0 \times 10^6\text{ J} - 2.0 \times 10^6\text{ J}\] \[10F = -10^6\] \[F = -10^5 \text{ N}\]

The negative sign indicates the force is antiparallel to the displacement.