Sign up for your FREE 7-Day trialGet instant access to all the benefits of SparkNotes PLUS! Cancel within the first 7 days and you won't be charged. We'll even send you a reminder.

SparkNotes Plus subscription is $4.99/month or $24.99/year as selected above. The free trial period is the first 7 days of your subscription. TO CANCEL YOUR SUBSCRIPTION AND AVOID BEING CHARGED, YOU MUST CANCEL BEFORE THE END OF THE FREE TRIAL PERIOD. You may cancel your subscription on your Subscription and Billing page or contact Customer Support at custserv@bn.com. Your subscription will continue automatically once the free trial period is over. Free trial is available to new customers only.

Step 2 of 4

Choose Your Plan

Step 3 of 4

Add Your Payment Details

Step 4 of 4

Payment Summary

Your Free Trial Starts Now!

For the next 7 days, you'll have access to awesome PLUS stuff like AP English test prep, No Fear Shakespeare translations and audio, a note-taking tool, personalized dashboard, & much more!

Thanks for creating a SparkNotes account! Continue to start your free trial.

Please wait while we process your payment

Your PLUS subscription has expired

We’d love to have you back! Renew your subscription to regain access to all of our exclusive, ad-free study tools.

Did you know you can highlight text to take a note?x

Problem :

A skier glides down a frictionless hill of 100 meters, the ascends another hill,
of height 90 meters, as shown in the figure below. What is the speed of the
skier when it reaches the top of the second hill?

The skier is in a conservative system, as the only force acting upon him is
gravity. Instead of calculating the work done over the curved hills, we can
construct an alternate path, because of the principle of path independence:

We construct a path of two segments: one is horizontal, going between the two
hills, and one is vertical, accounting for the vertical drop between the two
hills. What is the work done over each of these two segments? Since the
gravitational force is perpendicular to the displacement in the horizontal
segment, no work is done. For the second segment, the gravitational force is
constant and parallel to the displacement. Thus the work done is:
W = Fx = mgh = 10mg. By the Work-Energy Theorem, this net work causes an increase in
velocity. If the skier started with no initial velocity, then we can relate the
final velocity to the work done:

mv_{f}^{2} = 10mg

We can cancel the mass and solve for v_{f}
:

v_{f} = = 14m/s

Thus the final velocity of the skier is 14 m/s.

Problem :

What was the change in potential energy in the last problem, given that the mass
of the skier is 50 kg?

Remember that ΔU = - W. We had calculated that the gravitational force
exerted a work of 10mg during the entire trip. Thus the change in potential
energy is simply the negative of this quantity: ΔU = - 10mg = - 500g = - 4900
Joules. The potential energy lost is converted into kinetic energy, accounting
for the final velocity of the skier.

Problem :
What is the total energy of the mass-spring system shown below? The mass is
shown at its maximum displacement on the spring, 5 meters from the equilibrium
point.

Here we have a system of two conservative forces, mass and gravity. Even if
there are more than one conservative force acting in a system, it is still a
conservative system. Thus potential energy is defined, and we can calculate the
total energy of the system. Since this quantity is constant, we may choose any
position for the mass that we like. In order to avoid calculating kinetic
energy, we choose a point at which the mass has no velocity: at its maximum
displacement, the position shown in the figure above. Also, since energy is
relative, we may choose our origin to be the equilibrium point of the spring, as
shown in the figure. Thus both the gravitational force and the spring force
contribute to the potential energy: U_{G} = mgh = - 5mg = - 245 Joules. Also,
U_{s} = kx^{2} = (10)(5)^{2} = 125 Joules. Thus the total potential
energy, and hence the total energy is the sum of these two quantities:
E = U_{G} + U_{s} = - 120 Joules. Remember that answers may vary on this problem. If we
had chosen a different origin for our calculations, we would have gotten a
different answer. Once we have chosen an origin, however, the answer for total
energy must remain constant.

Problem :

A particle, under the influence of a conservative force, completes a circular
path. What can be said about the change in potential energy of the particle
after this journey?

We know that if the particle completes a closed path, the net work on the
particle is zero. We already established through the Work-Energy Theorem that
the total kinetic energy does not change. However, we also know that ΔU = - W. Since no work is done, the potential energy of the system does not change.

We can also answer this question in a more conceptual manner. We have defined
potential energy as the energy of configuration of a system. If our particle
returns to its initial position, the configuration of the system is the same,
and must have the same potential energy.

Problem :

A pendulum with string of length 1 m is raised to an angle of 30^{o} below
the horizontal, as shown below, and then released. What is the velocity of the
pendulum when it reaches the bottom of its swing?

In this case there are two forces acting on the ball: gravity and tension from
the spring. The tension, however, always acts perpendicular to the motion of the
ball, thus contributing no work to the system. Thus the system is a conservative
one, with the only work being done by gravity. When the pendulum is elevated, it
has a potential energy, according to its height above its lowest position. We
can calculate this height:

The height h can be calculated by subtracting x from the total length of the
string: h = 1 - x. We use a trigonometric relation to find x:
sin30^{o} = . Thus x = .5m and h = 1 - .5 = .5m. Now that we
have the initial height of the pendulum, we can calculate its gravitational
potential energy: U_{G} = mgh = .5mg. All of this potential energy is converted into
kinetic energy at the final position of the pendulum, with a height of 0. Thus:
.5mg = mv^{2}. The masses cancel, and we can solve for v:
v = = 3.1m/s. Thus, when the pendulum reaches an angle of 90 with the
horizontal, it has a velocity of 3.1 m/s.