20. yes, the ISS has both PE and KE. Its height is the altitude
30. The work required to penetrate the wood on the first strike of the hammer is proportional to the depth it penetrates, assuming a constant friction force per length along the side of the nail. With twice the hammer velocity, there is four times the kinetic energy transferred from hammer to nail.
KE2/KE1 = 4 = ((Force 2)/(Force 1) ([Depth2)/(Depth1)]
= 2 [(Depth2)/(Depth1)]
(Depth2)/(Depth1) = 4/2 = 2
This only applies to the first hit of the nail. After the nail is part way in, the friction force will depend upon the initial penetration length, and the penetration per strike will tend to vary with more closely with V^2.
20. does the international space station have gravitational pe? ke? explain?
-It has gravitation pe.
I thought that Gravitational potential energy is the energy stored in an object as the result of its vertical position or height?
30.a moving hammer hits a nail and drives it into a wall. if the hammer hits the nail with twice the speed, how much deeper will the nail be driven? if it hits with three times the speed?
work = velocity^2
force*distance=velocity^2
distance is proportional to velocity^2
so what is twice square?
would it be 4 times?
2 answers
The International Space Station, or any satellite, has both potential energy and kinetic energy.
Without getting into the derivation of the relationships, sufficeth to say that the the ability of a unit mass at a distance r to perform work, with respect to infinity, is Ep = - GM/r. This is called the potential energy of the unit mass in the gravitational field of mass M, at a distance r from the center of M. The kinetic energy, or energy of motion of the unit mass is Ek = mVo^2/2. If a mass m describes an elliptical path around the mass M, and the radius vector at a specific moment is r, the potential energy of the mass is Ep = -mGM/r. If the semi-major axis of the elliptical path is a, then the velocity at any point is given by V = sqrt[GM(2/r - 1/a)] and thus the kinetic energy is Ek = 1/2m(GM)(2/r - 1/a). The total energy of the mass is therefore Et = Ep + Ek = -GM/2a. Seeing that r has dropped out of the equation altogether tells us that the total energy of the mass is independent of the length of the radius vector. Therefore the the sum of the potential and kinetic energies of the mass remains constant over the elliptical orbit and as one increases the other decreases. At the perigee, the low point in the elliptical orbit, the potential energy is a minimum, and the kinetic energy (and therefore the velocity) is a maximum. At the apogee, the highest point in the orbit, we have the complete opposite, maximum potential energy and minimum kinetic energy. The potential energy is always negative and when the radius is equal to the semi-major axis distance, the point where the semi-minor axis intersects the ellipse, the absolute value of the kinetic energy is exactly half the potential energy. The greater the eccentricity of the ellipse, the greater are the variations of the energy. In the case of a circular orbit, eccentricity zero, the energies are constant and the kinetic energy is exactly half the potential energy.
Without getting into the derivation of the relationships, sufficeth to say that the the ability of a unit mass at a distance r to perform work, with respect to infinity, is Ep = - GM/r. This is called the potential energy of the unit mass in the gravitational field of mass M, at a distance r from the center of M. The kinetic energy, or energy of motion of the unit mass is Ek = mVo^2/2. If a mass m describes an elliptical path around the mass M, and the radius vector at a specific moment is r, the potential energy of the mass is Ep = -mGM/r. If the semi-major axis of the elliptical path is a, then the velocity at any point is given by V = sqrt[GM(2/r - 1/a)] and thus the kinetic energy is Ek = 1/2m(GM)(2/r - 1/a). The total energy of the mass is therefore Et = Ep + Ek = -GM/2a. Seeing that r has dropped out of the equation altogether tells us that the total energy of the mass is independent of the length of the radius vector. Therefore the the sum of the potential and kinetic energies of the mass remains constant over the elliptical orbit and as one increases the other decreases. At the perigee, the low point in the elliptical orbit, the potential energy is a minimum, and the kinetic energy (and therefore the velocity) is a maximum. At the apogee, the highest point in the orbit, we have the complete opposite, maximum potential energy and minimum kinetic energy. The potential energy is always negative and when the radius is equal to the semi-major axis distance, the point where the semi-minor axis intersects the ellipse, the absolute value of the kinetic energy is exactly half the potential energy. The greater the eccentricity of the ellipse, the greater are the variations of the energy. In the case of a circular orbit, eccentricity zero, the energies are constant and the kinetic energy is exactly half the potential energy.
0 answers