* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
PHYS 101 Lecture 10 - Work and kinetic energy 10 - 1 Lecture 10 - Work and Kinetic Energy What’s important: • impulse, work, kinetic energy, potential energy Demonstrations: • block on plane • balloon with propellor • conversion of work to P.E. Work In previous lectures, we investigated the effect of a force acting over a period of time. Newton’s Second Law is sometimes written in the form F = ma = m Δv / Δt ----> F Δt = m Δv = Δ(mv) = Δp impulse In other words, a force acting through time gives the impulse. What about a force acting through a distance? F Δx ≡ Work ≡ W (measured in Joules) Suppose the force produces a constant acceleration a. Then $ v 2 # v 2i ' $ v 2 # v 2i ' 1 1 2 2 W ! F • "x = F & f = ma & f = mvf # mvi % 2a ( % 2a ( 2 2 At this point, work begins to look like an interesting quantity because it depends only on the end - points (not on the path). Is it true in general? Yes, but only in the absence of friction (proof requires integration, can be found in PHYS 120 notes). We see that the work changes the kinetic energy 1/2 mv2 of the particle. In words, The work done by an unbalanced force is equal to the change in the kinetic energy of the object. As a problem-solving technique, construct a free-body diagram to determine the unbalanced force. In three dimensions, we must generalize F Δx. Since force and displacement are both vectors, then we expect W ≡ F • Δr. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 10 - Work and kinetic energy 10 - 2 Example A molecular motor can walk along a filament in a cell (actin or microtubule) carrying a cargo, or contracting a muscle: molecular motor The force it exerts is about 5 pN, and the power stroke is 5 nm. What is the work done in the power stroke? W = F • Δx = 5 x 10-12 x 5 x 10-9 = 2.5 x 10-20 Joules. This is about half the energy released during the hydrolysis of ATP. Example Find the work done by the centripetal acceleration for an object executing uniform circular motion. Δx F F = m a = -(mv2/r2) r ---> F ⊥ Δx ---> Work = 0 It’s no surprise that there is no work done, since there is no change in kinetic energy. When the forces are not constant, we must find the area under the force vs. distance curve: F W x We'll do an example of this for a spring in the following lecture. © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 10 - Work and kinetic energy 10 - 3 Potential Energy Consider the situation of an object sliding down a frictionless plane at an angle θ. ! L h The force due to gravity is mg, and it does work on the block W=F•L = mg • L sinθ = mgh. (using h = L sinθ) In turn, the work W results in a change in kinetic energy of the block: mvbottom2 /2 - mvtop2 /2 = mgh Suppose we now raise the block very, very slowly back up to its original position, by applying a force which just balances the force due to gravity. N Fexternal mg The total work is zero, since the forces cancel out in the direction of motion. Similarly, ΔK = 0 since the object is at rest at the top and bottom. The work we have done on the block has not resulted in a change in kinetic energy! Even though this is a nice consistent picture, we find it unsatisfactory because we are doing work on the object. In order to differentiate what we do as external agents, and what happens to the system in response to our work, we introduce the idea of potential energy. Here Wdone on the system = increase in potential energy U of the system In some sense, the work we have done is "stored" as potential energy, free to be converted into kinetic energy at some later time. The gravitational potential energy is then U = mgh. We can now write down a conservation law which states that © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 10 - Work and kinetic energy !E = 0 where E = U + K 10 - 4 (conservation of energy) U has replaced W or In the absence of dissipative forces [heat, friction...] and of forces for which it is not possible to define a potential energy, the total mechanical energy of a system is constant. [dissipative forces = friction, viscosity, etc.] Caveats: 1. Don’t double-count work and potential energy – use only one in a given problem. 2. Beware of the signs of work: Wdone on the system raises its potential energy Wdone by the system lowers its potential energy. For a particular situation, Wdone on the system = - Wdone by the system (somewhat like subtrction is the addition of a negative number). 3. The zero of the potential energy is arbitrary. Work only tells us the difference in potential, never its absolute magnitude. Example The work required to separate two bodies attracted by gravity from r = R to r = ∞, where r is the distance of separation, is ΔU = Gm1m2 / R. This work raises the gravitational potential energy of the objects. If we want to say U(r = ∞) = 0 [i.e. objects at infinite separation have no gravitational PE], then we set U = -Gm1m2 / R. The - sign should not bother us. The “zero” of the potential energy is not defined because we can only evaluate the difference in potential energy arising from the work Ugravity r We can relate this expression for the gravitational potential energy to the more familiar mgh by comparing U at the surface of the Earth Re with U at a distance Re+h: The two masses involved are the mass of the object m, and the mass of the Earth Me. First, we look at the force of gravity on the surface of the Earth, r = Re. According to © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited. PHYS 101 Lecture 10 - Work and kinetic energy 10 - 5 Newton's universal law of gravity, G Mem !# G Me $& F = = m = gm Re2 " Re2 % That is, in terms of the Earth's size and mass, G Me . g= Re2 Now, let's calculate the potential energy difference: ! U = [U at Re + h] " [U at Re ] =" = G Mem G Me m + Re + h Re G Me m #% Re &( 1" Re $ Re + h ' The term in the bracket can be approximated by Re R + h ! Re h h . 1! = e = " Re + h Re + h Re + h Re Then, the expression for ΔU becomes: G Mem h G Me !U " • = mh = gmh Re Re Re2 © 2001 by David Boal, Simon Fraser University. All rights reserved; further copying or resale is strictly prohibited.