
Document
... A force does positive work when it has a vector component in the same direction as the displacement, and it does negative work when it has a vector component in the opposite direction. It does zero work when it has no such vector component. ...
... A force does positive work when it has a vector component in the same direction as the displacement, and it does negative work when it has a vector component in the opposite direction. It does zero work when it has no such vector component. ...
Unit 5 Review Packet
... move the box. The ramp decreases the amount of distance the box must travel. The ramp decreases the amount of effort required to move the box. The ramp makes the box weigh less and require less force to move. ...
... move the box. The ramp decreases the amount of distance the box must travel. The ramp decreases the amount of effort required to move the box. The ramp makes the box weigh less and require less force to move. ...
Please tear off this top page carefully (only the top page!!!). The
... The balls then roll down the ramp without slipping (neglect air resistance). The two balls reach the bottom of the ramp at the same time. a. True b. False 2. A wheel is spinning clockwise and slowing down. The wheel’s angular velocity and angular acceleration vectors point in opposite directions. a. ...
... The balls then roll down the ramp without slipping (neglect air resistance). The two balls reach the bottom of the ramp at the same time. a. True b. False 2. A wheel is spinning clockwise and slowing down. The wheel’s angular velocity and angular acceleration vectors point in opposite directions. a. ...
Kinetic energy - GZ @ Science Class Online
... When sky divers reach terminal velocity they are traveling at a constant speed. The forces of gravity accelerating the skydiver towards earth are matched exactly by the force of friction from the air particles pushing against the skydiver. If the person wears a more aerodynamic suit or points their ...
... When sky divers reach terminal velocity they are traveling at a constant speed. The forces of gravity accelerating the skydiver towards earth are matched exactly by the force of friction from the air particles pushing against the skydiver. If the person wears a more aerodynamic suit or points their ...
TODAY:
... An object rotating about an axis tends to remain rotating about the same axis, unless an external influence (torque, see soon) is acting. (c.f. 1st law) The property to resist changes in rotational state of motion is called rotational inertia, or moment of inertia, I . Depends on mass, as well as th ...
... An object rotating about an axis tends to remain rotating about the same axis, unless an external influence (torque, see soon) is acting. (c.f. 1st law) The property to resist changes in rotational state of motion is called rotational inertia, or moment of inertia, I . Depends on mass, as well as th ...
Hooke`s Law and Potential Energy
... In this experiment we will neglect frictional terms and we will not do any work on the system between the two times we measure the energy. The energy of the system consists of the potential energy stored in the spring [Eq. (2)], the kinetic energy of the oscillating mass [KE = (1/2)mv 2], and the gr ...
... In this experiment we will neglect frictional terms and we will not do any work on the system between the two times we measure the energy. The energy of the system consists of the potential energy stored in the spring [Eq. (2)], the kinetic energy of the oscillating mass [KE = (1/2)mv 2], and the gr ...
Version B
... The rotational analog of linear momentum is angular momentum, L: Then the rotational analog of Newton’s second law is: ...
... The rotational analog of linear momentum is angular momentum, L: Then the rotational analog of Newton’s second law is: ...
Chapter 7
... IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for K 2 and then for v2 . SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel, as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by th ...
... IDENTIFY: Use the work-energy theorem, Eq.(7.7). Solve for K 2 and then for v2 . SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the barrel, as shown in Figure 7.53. Use F = kx to find the amount the spring is initially compressed by th ...
Chapter 07: Kinetic Energy and Work
... final kinetic energy = initial kinetic energy + net work ...
... final kinetic energy = initial kinetic energy + net work ...
Powerpoint
... • The Sun is further away from Earth than the Moon is • The change in the Sun’s gravitational force from one side of Earth to the other is quite small • The change in the Moon’s gravitational force from one side of the Earth to the other is quite ...
... • The Sun is further away from Earth than the Moon is • The change in the Sun’s gravitational force from one side of Earth to the other is quite small • The change in the Moon’s gravitational force from one side of the Earth to the other is quite ...
Instructor: Hacker Course: Physics 121
... This exam will consist of 21 multiple-choice problems. You may not use calculators or other electronic devices on this exam. The use of such a device will be regarded as an attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam are rough sketches: they are not gener ...
... This exam will consist of 21 multiple-choice problems. You may not use calculators or other electronic devices on this exam. The use of such a device will be regarded as an attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam are rough sketches: they are not gener ...
LinearMomentum - University of Colorado Boulder
... If the initial conditions (masses and initial velocities) are known, and we seek the final velocities, then we have two equations (Conserv of p, Conserv of KE) in two unknowns (vA' and vB' ), and it is possible to solve. But the algebra gets very messy, because of the squared terms in the KE ...
... If the initial conditions (masses and initial velocities) are known, and we seek the final velocities, then we have two equations (Conserv of p, Conserv of KE) in two unknowns (vA' and vB' ), and it is possible to solve. But the algebra gets very messy, because of the squared terms in the KE ...
Classical central-force problem
In classical mechanics, the central-force problem is to determine the motion of a particle under the influence of a single central force. A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center. In many important cases, the problem can be solved analytically, i.e., in terms of well-studied functions such as trigonometric functions.The solution of this problem is important to classical physics, since many naturally occurring forces are central. Examples include gravity and electromagnetism as described by Newton's law of universal gravitation and Coulomb's law, respectively. The problem is also important because some more complicated problems in classical physics (such as the two-body problem with forces along the line connecting the two bodies) can be reduced to a central-force problem. Finally, the solution to the central-force problem often makes a good initial approximation of the true motion, as in calculating the motion of the planets in the Solar System.