Work and energy

What is kinetic energy?

Learn what kinetic energy means and how it relates to work.

What is kinetic energy?

Kinetic energy is the energy an object has because of its motion.
If we want to accelerate an object, then we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed. The energy transferred is known as kinetic energy, and it depends on the mass and speed achieved.
Kinetic energy can be transferred between objects and transformed into other kinds of energy. For example, a flying squirrel might collide with a stationary chipmunk. Following the collision, some of the initial kinetic energy of the squirrel might have been transferred into the chipmunk or transformed to some other form of energy.

How can we calculate kinetic energy?

To calculate kinetic energy, we follow the reasoning outlined above and begin by finding the work done, WW, by a force, FF, in a simple example. Consider a box of mass mm being pushed through a distance dd along a surface by a force parallel to that surface. As we learned earlier
W=F⋅d=m⋅a⋅d\begin{aligned} W &= F \cdot d \\ &= m · a · d\end{aligned}
Huh? I'm lost already.
Work is defined to be W=Fd cosθW=Fd \text{ cos}\theta. If the force is parallel to the direction of motion, the angle is θ=0\theta=0, which makes cos 0=1\text{ cos } 0=1 and W=FdW=Fd.
Also, Newton's second law tells us that Fnet=maF_{net}=ma, so substituting into the previous formula we get W=madW=mad.
If we recall our kinematic equations of motion, we know that we can substitute the acceleration if we know the initial and final velocity—viv_\mathrm{i} and vfv_\mathrm{f}—as well as the distance.
What kinematic formula is this?
Rearranging the kinematic formula v2=v02+2adv^2=v_0^2+2ad gives us a=v2−v022da=\dfrac{v^2-v_{0}^2}{2d}.
W=m⋅d⋅vf2−vi22d=m⋅vf2−vi22=12⋅m⋅vf2−12⋅m⋅vi2\begin{aligned} W &= m\cdot d\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2d} \\ &= m\cdot \frac{v_\mathrm{f}^2-v_\mathrm{i}^2}{2} \\ &= \frac{1}{2}\cdot m \cdot v_\mathrm{f}^2 - \frac{1}{2}\cdot m \cdot v_\mathrm{i}^2 \end{aligned}
So, when a net amount of work is done on an object, the quantity 12mv2\dfrac{1}{2}mv^2—which we call kinetic energy KK—changes.
Kinetic Energy: K=12⋅m⋅v2\text{Kinetic Energy: } K=\frac{1}{2}\cdot m\cdot v^2
Alternatively, one can say that the change in kinetic energy is equal to the net work done on an object or system.
Wnet=ΔKW_{net}=\Delta K
This result is known as the work-energy theorem and applies quite generally, even with forces that vary in direction and magnitude. It is important in the study of conservation of energy and conservative forces.

What is interesting about kinetic energy?

There are a couple of interesting things about kinetic energy that we can see from the equation.
  • Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.
  • Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.
  • Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m/s.
Exercise 1a: Being in the wrong place when an African elephant—mass = 6000 kg, velocity = 10 m/s—is charging can really ruin your day. How fast would a 1 kg cannon ball travel if it had the same kinetic energy as the elephant?
Show solution.
Kcannon ball=KelephantK_{\text{cannon ball}}=K_{\text{elephant}}
12mballâ‹…v2=12melephantâ‹…velephant2\dfrac{1}{2}m_\mathrm{ball}\cdot v^2=\dfrac{1}{2}m_\mathrm{elephant}\cdot v_\mathrm{elephant}^2
vball=melephant⋅velephant2/mball=775 m/s(∼2790 km/hrv_{\text{ball}}=\sqrt{m_\mathrm{elephant}\cdot v_\mathrm{elephant}^2/m_\mathrm{ball}}=775~\mathrm{m/s} \quad (\sim 2790~\text{km/hr} or 1730 mi/hr)1730~\text{mi/hr})
Which is more than twice the speed of sound at sea level.
Exercise 1b: How would you expect the damage done to a brick wall to differ in the event of separate collisions with the elephant and cannonball?
Show solution.
Although both objects have the same kinetic energy, the damage done to an object is generally proportional to the energy transferred during the collision. The force due to collision with a cannonball is concentrated in a very small area. The ball might be expected to pass straight through the wall, making a neat hole and emerging from the other side with most of its velocity—and therefore kinetic energy—remaining. On the other hand, the collision with the elephant might result in force spread over a large area, and the whole wall might be knocked down.
Exercise 2: Hydrazine rocket propellant has an energy density EdE_d of 1.6MJkg1.6 \dfrac{\text{MJ}}{\text{kg}}. Suppose a 100 kg (mrm_r) rocket is loaded with 1000 kg (mpm_p) of hydrazine. What velocity could it achieve? To keep things simple, let’s assume that the propellant is burned up very quickly and that the rocket is not subject to any external forces.
Show solution.
The total chemical potential energy stored in the propellant is Edâ‹…mpE_\mathrm{d} \cdot m_\mathrm{p}. If we equate this to the kinetic energy, we get
12mrâ‹…v2=Edâ‹…mp\frac{1}{2} m_\mathrm{r} \cdot v^2 = E_\mathrm{d} \cdot m_\mathrm{p}
We can rearrange the equation to find the velocity vv:
v=2⋅Edmpmr=5657 m/sv = \sqrt{\frac{2\cdot E_\mathrm{d} m_\mathrm{p}}{m_\mathrm{r}}} = 5657 ~\mathrm{m/s}
Given that the velocity a spacecraft orbiting the earth is about 7 km/s, it would seem that we have gotten quite close to achieving orbit with this rocket. However, it turns out that in reality things are not quite so simple. One difference is that a real rocket does not burn all its propellant instantly, but at a given rate. It takes energy to lift the unburned fuel. Another factor is that a real rocket in the atmosphere loses energy to drag.
Nevertheless, the energy density of propellants and the ratio of the mass of the propellant to the mass of the vehicle are critical in rocketry. The practical limitations to these values is what makes orbit very difficult to achieve. Real rockets must use multiple stages which are discarded in flight to achieve orbital velocity. Discarding the mass of a lower stage allows the remaining stages to accelerate more easily to the final velocity.
Loading