First, we need to pick an origin for the
y -axis and then determine the value of the constant that makes the potential energy zero at the height of the base. Then, we can determine the potential energies from
[link] , based on the relationship between the zero potential energy height and the height at which the hiker is located.
Solution
Let’s choose the origin for the
y -axis at base height, where we also want the zero of potential energy to be. This choice makes the constant equal to zero and
$U\left(\text{base}\right)=U\left(0\right)=0.$
At the summit,
$y=147\phantom{\rule{0.2em}{0ex}}\text{m}$ , so
Besides illustrating the use of
[link] and
[link] , the values of gravitational potential energy we found are reasonable. The gravitational potential energy is higher at the summit than at the base, and lower at sea level than at the base. Gravity does work on you on your way up, too! It does negative work and not quite as much (in magnitude), as your muscles do. But it certainly does work. Similarly, your muscles do work on your way down, as negative work. The numerical values of the potential energies depend on the choice of zero of potential energy, but the physically meaningful differences of potential energy do not. [Note that since
[link] is a difference, the numerical values do not depend on the origin of coordinates.]
Check Your Understanding What are the values of the gravitational potential energy of the hiker at the base, summit, and sea level, with respect to a sea-level zero of potential energy?
In
Work , we saw that the work done by a perfectly elastic spring, in one dimension, depends only on the spring constant and the squares of the displacements from the unstretched position, as given in
[link] . This work involves only the properties of a Hooke’s law interaction and not the properties of real springs and whatever objects are attached to them. Therefore, we can define the difference of
elastic potential energy for a spring force as the negative of the work done by the spring force in this equation, before we consider systems that embody this type of force. Thus,
If the spring force is the only force acting, it is simplest to take the zero of potential energy at
$x=0$ , when the spring is at its unstretched length. Then, the constant is
[link] is zero. (Other choices may be more convenient if other forces are acting.)
Spring potential energy
A system contains a perfectly elastic spring, with an unstretched length of 20 cm and a spring constant of 4 N/cm. (a) How much elastic potential energy does the spring contribute when its length is 23 cm? (b) How much more potential energy does it contribute if its length increases to 26 cm?
Strategy
When the spring is at its unstretched length, it contributes nothing to the potential energy of the system, so we can use
[link] with the constant equal to zero. The value of
x is the length minus the unstretched length. When the spring is expanded, the spring’s displacement or difference between its relaxed length and stretched length should be used for the
x -value in calculating the potential energy of the spring.
Solution
The displacement of the spring is
$x=23\phantom{\rule{0.2em}{0ex}}\text{cm}-20\phantom{\rule{0.2em}{0ex}}\text{cm}=3\phantom{\rule{0.2em}{0ex}}\text{cm}$ , so the contributed potential energy is
$U=\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(4\phantom{\rule{0.2em}{0ex}}\text{N/cm}\right){\left(3\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}=0.18\phantom{\rule{0.2em}{0ex}}\text{J}$ .
When the spring’s displacement is
$x=26\phantom{\rule{0.2em}{0ex}}\text{cm}-20\phantom{\rule{0.2em}{0ex}}\text{cm}=6\phantom{\rule{0.2em}{0ex}}\text{cm}$ , the potential energy is
$U=\frac{1}{2}k{x}^{2}=\frac{1}{2}\left(4\phantom{\rule{0.2em}{0ex}}\text{N/cm}\right){\left(6\phantom{\rule{0.2em}{0ex}}\text{cm}\right)}^{2}=0.72\phantom{\rule{0.2em}{0ex}}\text{J}$ , which is a 0.54-J increase over the amount in part (a).
Significance
Calculating the elastic potential energy and potential energy differences from
[link] involves solving for the potential energies based on the given lengths of the spring. Since
U depends on
${x}^{2}$ , the potential energy for a compression (negative
x ) is the same as for an extension of equal magnitude.
A 10kg ball travelling at 4meter per second collides elastically in a head-on collision with a 2kg ball.What are (a)the velocities and (b)the total momentum of the balls after collision?
a)v1 8/3s&v2 20/3s. b)in elastic collision total momentum is conserved.
Bala
The displacement of the air molecules in sound wave is modeled with the wave function s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t)s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t) . (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmon
If the block is displaced to a position y , the net force becomes Fnet=k(y−y0)−mg=0Fnet=k(y−y0)−mg=0 . But we found that at the equilibrium position, mg=kΔy=ky0−ky1mg=kΔy=ky0−ky1 . Substituting for the weight in the equation yields. Show me an equation of graph.
Is equal to the square of the velocity divided by the radius of circular path of the object
Mukhtaar
how to find maximum acceleration and velocity of simple harmonic motion?
chander
how to find maximum acceleration and velocity of simple harmonic motion and where it occurres?
chander
you can use either motion equations or kinetic equation and potential equation .
lasitha
how destraction 1kg uranium
Sayed
A Radial Acceleration is defined as the upward movement of an object.
Andrew
A body of 2.0kg mass makes an elastic collision with another at rest and continues to more in the original direction but with 1/4 of its ori is the mass of the struck body?
I believe because speed is a function of air density, and colder air is more dense
Jerry
At night air is denser because of humidity.
Clifton
Night air is cooler. Sound requires medium to travel so the denser the medium the fastest the sound travels. Humid air is denser then warmer air as in day.
Clifton
The humidity statement is misleading , colder air is more dense period.
Jerry
because there is no any other sound to reverberate with it so it clearly travel to lot of distance
and also humidity
and also due to denser air at night
Azam
please could you guys help me with physics best websites
Baje
because it is quiet at night. this takes us to the topic wave, it depends on the wave at that moment, which Echo's....sound travelled.
Andrew
because it is quite at night. this takes us to the topic wave , it depends on the wave at that the physics