PHY 107 1st Edition Lecture 13 Outline of Last Lecture I. Drag Force & Terminal SpeedII. Uniform Circular Motion Outline of Current LectureIII. Kinetic EnergyIV. WorkCurrent LetureKinetic Energy: K – describes the state of motion of an object of mass m and speed v- K = mv22 (unit is Joules) Work: W – If a force is applied to an object of mass m, the object can either accelerate and increase its speed, v, and kinetic energy, K, or it can decelerate the object and decrease its kinetic energy. - W = Fdcosθ W = ´F · ´d- We account for these changes in K by saying that F has transferred energy W to or from the object - If energy is transferred to m (its K increases), we say that work was done by F on the object (W>0) - If energy is transferred from the object (its K decreases) we say that work was done by m (W>0) - Ex. A bead is moving along a straight wire on the x-axis, ´F is applied at an angle of θ to the wire These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.o Ki = mv022 and Kf = mv22 → change in K: Kf – Ki = Fdcosθo Work done: W = Fxd = Fdcosθ - Note 1: the expression for work only applies when F is constant- Note 2: we assume that the moving object is point like- Note 3: W > 0 if 0 < θ < 90° and W < 0 if 90° < θ < 180°- Net Work: If we have several forces acting on a body, there are two methods that can be used to calculate the net work (Wnet):o Method 1: First calculate the work done by each force: ex. WA by force WB by force ´FB, and WC by force ´FC, then determine Wnet = WA + WB + WCo Method 2: Calculate first ´Fnet = ´FA + ´FB + ´FC,then determine Wnet = ´Fnet · ´dWork – Kinetic Energy Theorem:- Alternative approach to mechanics – use scalars instead of vectors- Kf – Ki = Wnet and ∆K = Kf – Ki so ∆K = Kf – Ki = Wnet↓ Change in kinetic energy of a particle = work done on a particle- If Wnet > 0 (0 < θ < 90°) → Kf – Ki > 0 → Kf > Ki- If Wnet < 0 (90° < θ < 180°) → Kf – Ki < 0 → Kf < KiWork done by gravitational force: - Throwing an object up and letting it fall back down- As it moves upward, Wg = mgdcos180° = -mgd- As it falls back down, Wg = mgdcos0° = mgdWork done by lifting an object:- An object of mass m is lifted by a force F from A to B, starting at rest at A and arriving at B with zero speed and force ´F is not necessarily constant during the trip - ∆K = Kf – Ki = Wnet → Ki = Kf → ∆K = 0 → Wnet = 0- Two forces: gravitational force and the force lifting the object - Wnet = Wa(A → B) + Wg(A → B) = 0 → Wa(A → B) = - Wg(A → B) - Wg(A → B) = mgdcos180° = -mgd → Wa(A → B) = mgdWork done by lowering an object: - The object mentioned above moves from B to A- Wg(B → A) = mgdcos0° = mgd, Wa(B → A) =
View Full Document