10.2: Work and energy
Imagine now that I’m picking up my 7 kg cat from section 10.1. I need to exert some force to do so, and I increase his potential energy—the work I did on him was positive. Now consider a different situation. Let’s say there is a car whose parking brake failed, and you (being foolish and heroic) jump in front of the car, push against it, and stop it. As you push on the car, it slows down: you are reducing the car’s kinetic energy. You are doing negative work on the car. So, despite being a scalar quantity, work can be negative:
- Work is positive when the applied force is acting in the same direction as the displacement.
- Work is negative when the applied force is acting in the opposite direction as the displacement.
As explained earlier, work is the relationship that connects force and energy. When energy is dissipated from a system, which we referred to as “waste” in chapter 6, the energy dissipated is equivalent to the work done by the dissipating force. Recall that the “waste” we worked with in chapter 6 always had a negative value, representing energy leaving the system. When a force is exerted in the opposite direction as an object’s motion, the work done by the force is negative.
The common example we consider in this book is energy dissipation due to the force of friction. As an object slides across a surface, friction does work on the object. The energy dissipated is the work done by friction. The work done by friction is negative because friction always acts on an object in the opposite direction than the object is moving.
Now we know the connection between force and energy, we can apply this to problems where we consider conservation of energy.
EXAMPLE 10.3
A cat is riding a skateboard on a smooth horizontal surface, traveling at 2 m/s. She rolls through a shallow puddle that is 0.5 m long, and is slowed down to 1.5 m/s. What is the magnitude of the force that the water in the puddle applied to the skateboard? The combined mass of the cat and skateboard is 10 kg.
Consider energy conservation:
\[
\begin{align*}
W &= E_f – E_i \\
-Fd &= K_f – K_i \\
&= \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2 \\
\hookrightarrow F &= -\frac{m(v_f^2 – v_i^2)}{2d} \\
&= -\frac{(10\ \textrm{kg})\left((1.5\ \textrm{m/s})^2 – (2\ \textrm{m/s})^2\right)}{2(0.5\ \textrm{m})} \\
&= 17.5\ \textrm{N}
\end{align*}
\]
Note that I used a negative sign for the work \(\left(W = -Fd\right)\); this is because the energy is leaving the system as the cat slows down.
EXAMPLE 10.4
A box full of snakes \(m = 6\ \textrm{kg}\) is resting on a frictionless level surface 0.5 m above the ground. It is given a push so that it has an initial speed of 1 m/s. After sliding down a frictionless ramp, the box goes across a level surface where is a frictional force of 17.64 N between the box and the surface. How long does this surface need to be to reduce the box to half of its initial speed?
This is similar to the previous problem, but now we have gravitational potential energy as well as kinetic energy playing a part in the initial energy of the box.
\[
\begin{align*}
W &= E_f – E_i \\
-F^fd &= K_f – (K_i + U_G) \\
-F^fd &= \frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2 – mgh \\
F^fd &= mgh + \frac{1}{2}mv_i^2 – \frac{1}{2}mv_f^2 \\
\hookrightarrow d &= \frac{mgh}{F^f} + \frac{1}{2}\frac{m(v_i^2 – v_f^2)}{F^f} \\
&= \frac{(6\ \textrm{kg})\left(9.81\ \textrm{m/s}^2\right)(0.5\ \textrm{m})}{17.64\ \textrm{N}} + \frac{1}{2}\frac{(6\ \textrm{kg})\left((1\ \textrm{m/s})^2 – (0.5\ \textrm{m/s})^2\right)}{17.64\ \textrm{N}} \\
&= 1.8\ \textrm{m}
\end{align*}
\]
EXAMPLE 10.5
A 600 g block is moving with a velocity of 4 m/s when it hits a massless spring \(\left(k = 2940\ \textrm{N/m}\right)\). Under the spring, there is a 700 N frictional force between the block and the ground. How far will the spring compress?
\[
\begin{align*}
W &= E_f – E_i\\
-F^fd &= U_f – K_i \\
-F^fx &= \frac{1}{2}kx^2 – \frac{1}{2}mv_i^2
\end{align*}
\]
This is quadratic in \(x\). We will solve using the quadratic formula.
\[
\begin{align*}
0 &= \underbrace{\frac{1}{2}k}_{a} x^2 + \underbrace{F^f}_{b}x \underbrace{-\frac{1}{2}mv_i^2}_{c} \\
\hookrightarrow x &= \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\
&= \frac{-F^f \pm \sqrt{\left(F^f\right)^2 – 4\left( \frac{1}{2}k\ \right)\left( -\frac{1}{2}mv_i^2 \right)}}{2\left( \frac{1}{2}k \right)} \\
&= \frac{-F^f \pm \sqrt{\left(F^f\right)^2 + kmv_i^2}}{k} \\
&= \frac{-(700\ \textrm{N}) \pm \sqrt{(700\ \textrm{N})^2 + (2940\ \textrm{N/m})(0.600\ \textrm{kg})(4\ \textrm{m/s})^2}}{2940\ \textrm{N/m}} \\
&= 0.68\ \textrm{cm}
\end{align*}
\]
There are two solutions to a quadratic equation. In this case, we used the positive solution; the variable \(x\) measures the absolute distance the spring compresses, so the negative solution was not physically meaningful.