Work and Energy in Physics
These are my notes on work-energy problems in physics.
Work-energy analysis is a very powerful tool for solving problems. Force
analysis works well in problems where there is a constant or where there is a
balance of forces.
But when the force acts over a distance, especially when that force is not
constant, then work-energy analysis is the correct analytic tool. To handle the
transformation of force over a distance, we first need to define the concepts of
kinetic and potential energy.
Kinetic Energy
A constant force applied to a mass over a distance produces an acceleration
according to F=ma. This acceleration over a distance changes the velocity of the
mass in accord with the kinetic equation \(v^2=v^2_{o}+2a(x-x_{o})\). If an in
this equation is replaced with F/m, then \(v^2=v^2_{o}+2(F/m)(x-x_{o})\). If the
constant force over a distance is identified as the work, then:
\[F(x-x_{o})=W=mv^2/2-mv^2_{o}/2\]
This simple derivation suggests a more formal look at the work performed on a
mass by a force.
The work performed on a mass by a force is defined as the integral of the force
times the distance over a specified distance. Mathematically, this is \(\int
F*ds\). If the force is replaced by ma and the distance taken in one dimension,
then the work integral becomes:
\[W=\int_{x1}^{x2} madx\]
The acceleration can be written as a chain derivative:
\[a=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\]
Then:
\[adx=\frac{dy}{dx}\frac{dx}{dt}dx=\frac{dx}{dt}dv=vdv\]
Which makes the integral:
\[W=\int_{v1}^{v2} mvdv=\frac{mv_{2}^{2}}{2}-\frac{mv_{1}^{2}}{2}\]
When a force is applied to a mass over a distance and that mass accelerates,
then there is a difference in velocity of the mass between before and after the
application of the force. The work performed is manifest in this velocity, and
the \(mv^2/2\) form is called the kinetic energy. Work is transformed into
kinetic energy.
Gravitational Potential Energy
Gravitational potential energy comes from the acceleration due to gravity and
the accompanying force on a mass on the surface of the Earth. If a mass is
raised from a height\( y_{1}\), to a height \(y_{2}\), directly opposite the
acceleration due to gravity, then work performed appears as gravitational
potential energy.
\[W=\int F*ds=\int_{y1}^{y2} mgdy=mgy_{2}-mgy_{1}\]
This mgy term is the gravitational potential. In mechanical systems, the total
energy of the system is the sum of the kinetic plus potential energies. In a
mechanical system on the surface of the Earth, work performed on the system can
appear as kinetic or potential energy.
work performed = change in kinetic energy + change in potential energy
A convenient way of understanding this is to say that work goes into kinetic or
potential energy.
Look first at a block sliding down a frictionless inclined plane. The velocity
at the bottom of the plane is related to the height h from which the block fell.
The kinetic energy at the bottom of the plane is \(mv^2/2\), and this must equal
mgh, the potential energy at the top of the plane. The gravitational potential
energy goes into kinetic energy.
\[\frac{mv^2}{2}=mgh\]
If mass =3.0 kg
Inclined plane =4.0 m
Angle at bottom of inclined plane and ground is 58 degrees
Height =3.4 m
We get:
\[v=\sqrt{2(9.8m/s^2)3.4m}=8.2m/s\]
If the block were to continue to slide horizontally along a frictionless
surface, it would maintain this 8.2 m/s velocity.
Frictional Forces
Frictional forces are proportional to the normal force(the force between two
surfaces) and a constant characteristic of the interface(the surfaces involved).
They also act to oppose the motion. Frictional forces acting over a distance
result in energy lost due to friction.
A coefficient of friction between the block and a flat surface at the bottom of
the plane would result in energy lost to friction and the block eventually
coming to rest. In this problem, the potential energy at the top of the incline
plane goes into kinetic energy at the bottom of the plane, and this energy goes
into work to overcome friction.
Give the flat surface a coefficient of 0.20, and the frictional retarding force
is:
\[f=\mu mg=0.20(3kg)9.8m/s^2=5.9N\]
To find the distance the block slides, we can go back to the kinetic energy at
the bottom of the plane or the potential energy at the top of the plane. Since
there are no lossed, thes enumbers must be the same. Take the potential energy
at the top of the plane because it is easier to calculate and is original data
in the problem. This energy must equal the work due to friction fl or:
\[mgh=fL=\mu mgL\]
\[3.0kg(9.8m/s^2)3.4m=100J=0.20(3kg)(9.8m/s^2)L\]
\[L=17m\]
Example 1
Place a 3 kg block at the top of a 3.4 m high frictionless incline. At the
bottom of the incline, the block encounters a spring with a constant of 400 N/m
on a horizontal surface. No energy is lost to friction. how far is the spring
compressed?
Solution:
The energy at the top of the plane, which is the same as the energy at the
bottom of the plane, goes into compressing the spring. The energy at the top of
the plane is:
\[100J=\frac{kx^2}{2}\]
\[x=\frac{200J*m}{400N}^{1/2}=0.71m\]
Example 2
Now complicate the first problem by adding a coefficient of friction of 0.20 for
the horizontal surface. How far does the block slide while compressing the
spring?
Solution:
Now the potential energy goes into compressing the spring and overcoming
friction. This statement is very helpful in writing the equation. In words, the
potential energy mgh equals the energy to compress the spring \(kx^2/2\) plus
the work to overcome friction \(\mu mgx\) or:
\[mgh=\frac{kx^2}{2}+\mu mgx\]
\[100J=(200N/m)x^2+0.20(3kg)(9.8m/s^2)x\]
\[200x^2+5.9x-100=0\]
\[x=\frac{-5.9 \pm \sqrt{(5.9)^2-4(200)(-100)}}{2(200)}=\frac{-5.9 \pm
283}{400}=0.70m\]
The negative root will not work for this problem, it does not make sense. The
block compresses the spring 0.70 m while sliding on this frictional surface.
It is instructive to do a problem first with force analysis and then with
work-energy analysis. Consider the case of a block sliding down an inclined
plane with a coefficient of friction. As will become evident, it is possible to
do this problem with force analysis techniques. Work-energy analysis is
conceptually and computationally easier.
Example 3
Consider the inclined plane with friction. Calculate the velocity of the block
at the bottom of the plane using force analysis and then work-energy analysis.
Solution:
The unbalanced force of 20 N acts on the 5 kg block, causing it to accelerate at 4 m/s^2 down the plane. The slant height of the plane is 6m*sin35 degrees = 10.5 m.
This acceleration over the 10.5 m results in a velocity of:
\[v^2=2a(x-x_{o})=2(4m/s^2)10.5m=9.2m/s\]
Solution2:
Now do the same using work-energy analysis. When the block is at the top of the
plane, the energy is:
\[mgh=5kg(9.8m/s^2)(6m=294J\]
At the bottom of the plane, all this potential energy has gone into kinetic
energy(velocity of the block) except for the amount used to do work against the
frictional force. The work(against the frictional force) is:
\[\mu NL=0.20(40N)10.5=84J\]
The energy left after the block has slid down the plane is the 294 J at the top
of the plane minus the 84 J lost to friction, and this energy(210J) is manifest
in the velocity of the block(kinetic energy). This energy is equal to the
\(mv^2/2\), the kinetic energy of the block at the bottom of the plane.
\[\frac{(5kg)v^2}{2}=210J=9.2m/s\]
Example 4
A 3 kg block sliding at 12 m/s along a frictionless surface encounters a spring
with a force constant of 500 N/m resting on a surface with a coefficient of
friction of 0.25. The friction surface is only under the spring. What is the
maximum compression of the spring?
Solution:
When the block encounters the spring and the friction surface, all the energy is
kinetic, \(mv^2/2\). This energy goes into compressing the spring and sliding
along the friction surface. The equation looks like this:
\[\frac{mv^2}{2}+\frac{kx^2}{2}+\mu mgx\]
be careful in writing this statement. Many people write the kinetic energy
correctly, the energy to compress the spring correctly, and then write the force
to slide along the friction surface. Be aware of this mistake, and check the
units to be sure that the units on one side of the equation are the same as the
units on the other side of the equation.
You may have noticed that in all the problems the first equation is written with
units just as a reminder that checking units will reduce the number of errors.
It is very frustrating to make a simple mistake that could have been avoided had
units been checked in the problem before proceeding with the numerical
calculations.
\[\frac{3kg(12m/s)^2}{2}=\frac{500N}{m}\frac{x^2}{2}+0.25(3kg)(9.8m/s^2)x\]
The units are correct, so without units the equation is:
\[250x^2+7.35x-216=0\]
\[x=\frac{-7.35 \pm \sqrt{7.35^2-4(250)(-216)}}{2*250}=\frac{-7.4 \pm
464.8}{500}=0.91m\]
This is the amount the spring is compressed as the block comes to a stop.
Example 5
For the situation described previously, find the velocity with which the block
leaves the spring.
Solution:
Start this problem with the spring compresses the 0.91 m. At this point the
potential energy stored in the spring is used to push the block 0.91 m back
across the friction surface. The energy remaining as the block leaves the spring
is the original kinetic energy less the energy expended in two 0.91 trips across
this friction surface. In word equation form, this is:
\[KE_{initial}-2(\mu mgx's)=KE_{final}\]
The total work performed in the block moving across the surface is:
\[2[0.25(3kg)(9.8m/s^2)0.91m]=13.4J\]
The velocity of the block as it flies off the spring is:
\[\frac{3kg(144m/s^2)}{2}-13.4J=\frac{(3.0kg)v^2_{f}}{2}=v_{f}=11.6m/s\]
Example 6
Go back to an inclined plane problem but with the addition of a spring with a
force constant of 300 N/m placed at the bottom of the incline. Calculate the
compression of the spring when the block slides down the plane.
Solution:
An approximate answer can be obtained by setting mgh, where h is the vertical
distance from the block to the spring, equal to \(kx^2/2\), then compresison of
the spring then becomes:
\[2.5kg(9.8m/s^2)4m=\frac{300N}{m}\frac{x^2}{2} \rightarrow x=0.81m\]
The compression of the spring gives the block a little more potential energy, so
a more correct statement would take into account the additional height h.
\[mgh+(mgx)\sin40=\frac{kx^2}{2}\]
\[h'=x(\sin40)\]
\[2.5kg(9.8m/s^2)4m+2.5kg(9.8m/s^2)x(\sin40)=\frac{300N}{m}\frac{x^2}{2}\]
Eliminating the units and cleaning up the numbers yields a quadratic.
\[150x^2-15.7x-98=0\]
\[x=\frac{15.7 \pm \sqrt{15.7^2-4(150)(-98)}}{2(150)}=\frac{15.7 \pm 243}{300}
=0.86m\]
This value is larger than what was calculated earlier due to the additional
height.
The procedure of finding an approximate solution and then refining it to the
exact solution is very helpful. First, you get an approximate solution that is
close to the exact one, and second, you educate yourself as to how to proceed to
the exact solution.
Example 7
Place a mass on a track made up of a flat section L with coefficient of friction
\(\mu\) and two frictionless semicircular surfaces of radius R. Let the mass
start from the top of one of the semicircles and calculate where it comes to
rest.
Solution:
The initial potential energy is mgR. When the mass encounters the friction
surface, this potential energy is dissipated in doing work to overcome friction.
Assuming that the energy lost due to friction in one traverse is less than the
initial potential energy, the mass will rise to a height(on the opposite
semicircle) R' dictated by the energy statement:
\[mgR'=mgR-\mu mgL\]
After another traverse of the flat portion of the track, the height will be
dictated by:
\[mgR''=mgR-2 \mu gL\]
and so on until all the original potential energy is dissipated.
Example 8
For the track in the previous example, take \(\mu =0.10\), R=1, and L=2m, and
find where the mass stops.
Solution:
The only way the mass loses energy is by sliding along the flat friction part of
the track. The initial potential energy goes into frictional work.
\[(mg)1.0m=0.10(mgx) \rightarrow x=10m\]
The mass crosses the friction area five times and ends up at the edge of the
friction surface opposite from where it started.
Example 9
Consider a 10g bullet passing through a 3kg block resting on a table. The
velocity of the bullet is 400 m/s on entering the block and 250 m/s on exit.
Calculate the energy lost by the bullet in passing through the block.
Solution:
The energy is all kinetic, so calculating before and after gives;
\[KE_{i}=(1/2)(10*10^{-3}kg)(400m/s^2)=800J\]
\[KE_{f}=(1/2)(10*10^{-3}kg)(250m/s^2)=312J\]
The kinetic energy lost = 488J
Example 10
If 30 percent of the energy lost in the previous example is available to move
the block along a surface with a coefficient of friction of 0.80, how far will
it move?
Solution:
Equate the energy available to the work performed.
\[0.30\Delta KE=\mu mgL\]
\[0.30(488J)=0.80(3kg)(9.8m/s^2)L \rightarrow L=6.2m\]
Example 11
Consider an elevator weighing 4000 N held 5 m above a spring with a force
constant of 8000 N/m. The elevator falls onto the spring while subject to a
frictional retarding force(brake) of 1000 N. Describe the motion of the
elevator.
Solution:
The potential energy of the elevator with respect to the top of the spring is:
\[mgh=4000N(5m)=20000J\]
When the elevator falls, this energy, less the energy lost due to the frictional
brake, is available to compress the spring.
\[20000J-1000N(5m)=15000J=energy-to-compress-spring\]
The friction brake stays on while the elevator is in contact with the spring, so
this energy goes into \(kx^2/2\). In equation form, this is:
\[15000J=k(x^2/2)+fx\]
This expression is only approximately correct because if the spring is
compressed a distance x, then this mgx provides an additional amount of
potential energy. The correct equation that determines the compression of the
spring is:
\[mg(h+x)=kx^2/2+f(h+x)\]
Putting numbers into the equation:
\[(4000N/m)x^2+(1000N-4000N)x+(1000N-4000N)5m=0\]
The units are correct so \(4000x^2-3000x-15000=0\)
\[4x^2-3x-15=0\]
\[x=\frac{3 \pm \sqrt{9-4(4)(-15)}}{2(4)}=2.3m\]
The energy stored in the spring at this compressed point is:
\[kx^2/2=8000N/m\frac{2.3m^2}{2}=22040J\]
This amount of energy will raise the elevator and do work against friction
according to the equation:
\[22040J=mgx'+fx'\]
\[22040J=(5000N)x'\]
\[x'=4.4m\]
This places the elevator 2.1 m above the spring.
This analysis can be repeated until the elevator comes to a stop.