Chapter C7 More on potential energy functions Do problems C7.B2,B3, C7.S1 Due Monday. Printouts with all your grades to present are in your folder. Please check them and report to me anything on the printout that is incorrect.

A note for lab. The equation we have for total energy was E =K 1 +K 2 +K 3 +…V(r 12 )+ V(r 13 )+ V(r 23 )+… It is possible to have more than one type of kinetic energy. K i represents the energy due to the center of mass, we can also have rotational kinetic energy. The equations for this are given in the lab.

Electromagnetic interaction where k=8.99x10 9 j·m/C 2 q 1 and q 2 are charges separated by a distance r It is important to begin a formula sheet that will continue throughout the semester. A good formula sheet will be very helpful on the problems and tests.

Case for like and unlike charges q 1 q 2 >0 V(r) q 1 q 2 <0 V(r) RepulsiveAttractive Like charges repel Unlike charges attract One charge is at the origin One way to remember if the potential is attractive or repulsive is to consider the direction a ball would roll if placed on the potential.

Gravitational attraction between two particles G = 6.67 x j·m/kg 2 m 1 and m 2 are the masses of two particles a distance r apart. Near the surface of the Earth this becomes V(z)=mgz V(r) The gravitational force is always attractive. This is a straight line close to the Earth’s surface because it represents a very small portion of the curve above Surface of the Earth

Comparison of electromagnetic and gravitational potentials. G = 6.67 x j·m/kg 2 k=8.99x10 9 j·m/C 2 The electromagnetic potential is times greater than gravity.

Potential energy of a spring V(x)=½ k s x 2 where x is the amount the spring is compressed or stretched k s is the spring constant, tells how hard it is to stretch the spring. F=-k s x where F is the force to stretch the spring and x is the distance it is stretched.

What must the spring constant “k” of a spring be so that when the spring is compressed 55 cm and 55kg person sits on the spring, it shoots them 2 m into the air? Data  v i =0, v f =0, z i =0, z f =2 m (g=9.8m/s 2 ) Principle – because the Earth floats in space, energy is conserved. The spring potential energy will be transformed into gravitational potential energy. We choose the zero level of potential energy to be the position of the person when the spring is compressed (and they are sitting on top of it)

What must the spring constant “k” of a spring be so that when the spring is compressed 55 cm and 55kg person sits on the spring, it shoots them 2 m into the air? spring P.E. = gravitational P.E. Principle: The spring potential energy is transformed into gravitational potential energy as the person is shot into the air. k=7100 N/m This corresponds to about 40 lbs/in h is the distance the spring is compressed. z f is the height the person rises

How fast must a rocket be moving to escape the Earth’s gravity? The radius of the Earth is 6380 km. What is the principle needed to solve the problem? Once the rocket obtains the escape speed it has kinetic energy that will be transformed into potential as it moves away from Earth. We need to move the rocket out of the potential well K.E.  P.E V(r)

Because the Earth is isolated in space, energy is conserved and we can write: Change in KE of rocket Change in KE of the Earth = 0 because the Earth is so massive. Potential energy r f →∞, v f →0 What is v i ? V i = 11.2 km/s There are two objects, the Earth and the rocket, and we must consider the interaction between these two objects.

Problems for this chapter Do problems C7.B2,B3 C7.S1 Due Monday.