Chapter 7: Kinetic Energy and Work
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Chapter 7: Kinetic Energy and Work
This approach is alternative approach to mechanics. It uses scalars such as work and kinetic energy rather than vectors such as velocity and acceleration. Therefore it simpler to apply.
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Nội dung Text: Chapter 7: Kinetic Energy and Work
 Chapter 7 Kinetic Energy and Work In this chapter we will introduce the following concepts: Kinetic energy of a moving object Work done by a force Power In addition we will develop the workkinetic energy theorem and apply it to solve a variety of problems This approach is alternative approach to mechanics. It uses scalars such as work and kinetic energy rather than vectors such as velocity and acceleration. Therefore it simpler to apply. (71)
 Kinetic Energy: We define a new physical parameter to describe m m the state of motion of an object of mass m and speed v mv 2 We define its kinetic energy K as: K= 2 We can use the equation above to define the SI unit for work (the joule, symbol: J ). An object of mass m = 1kg that moves with speed v = 1 m/s has a kinetic energy K = 1J Work: (symbol W) If a force F is applied to an object of mass m it can accelerate it and increase its speed v and kinetic energy K. Similarly F can decelerate m and decrease its kinetic energy. We account for these changes in K by saying that F has transferred energy W to or from the object. If energy it transferred to m (its K increases) we say that work was done by F on the object (W > 0). If on the other hand. If on the other hand energy its transferred from the object (its K decreases) we say that work was done by m (W < 0) (72)
 Finding an expression for Work: m m Consider a bead of mass m that can move without friction along a straight wire along r the xaxis. A constant force F applied at an angle φ to the wire is acting on the bead We apply Newton's second law: Fx = max We assume that the bead had an initial r r r velocity vo and after it has travelled a distance d its velocity is v . We apply the third equation of kinematics: v 2 − vo = 2ax d We multiply both sides by m / 2 → 2 m 2 m 2 m m F m 2 v − vo = 2ax d = 2 x d = Fx d = F cos φ d K i = vo 2 2 2 2 m 2 m 2 K f = v → The change in kinetic energy K f − K i = Fd cos ϕ 2 Thus the work W done by the force on the bead is given by: W = Fx d = Fd cos ϕ r r W = Fd cos ϕ W = F ⋅d (73)
 r (74) FA r W = Fd cos ϕ m m FC r r W = F ⋅d r FB The unit of W is the same as that of K i.e. joules Note 1:The expressions for work we have developed apply when F is constant Note 2:We have made the implicit assumption that the moving object is pointlike Note 3: W > 0 if 0 < φ < 90° , W < 0 if 90° < φ < 180° Net Work: If we have several forces acting on a body (say three as in the picture) there are two methods that can be used to calculate the net work Wnet r Method 1: First calculate the work done by each force: WA by force FA , r r WB by force FB , and WC by force FC . Then determine Wnet = WA + WB + WC r r r r r r Method 2: Calculate first Fnet = FA + FB + FC ; Then determine Wnet = F ⋅ d
 WorkKinetic Energy Theorem We have seen earlier that: K f − K i = Wnet . m m We define the change in kinetic energy as: ∆K = K f − K i . The equation above becomes the workkinetic energy theorem ∆K = K f − K i = Wnet Change in the kinetic net work done on energy of a pareticle = the particle The workkinetic energy theorem holds for both positive and negative values of Wnet If Wnet > 0 → K f − K i > 0 → K f > K i If Wnet < 0 → K f − K i < 0 → K f < K i (75)
 B Work Done by the Gravitational Force: Consider a tomato of mass m that is thrown upwards at point A with initial speed vo . As the tomato rises, it slows down by the A gravitational force Fg so that at point B its has a smaller speed v. The work Wg ( A → B ) done by the gravitational force on the tomato as it travels from point A to point B is: Wg ( A → B ) = mgd cos180° = −mgd The work Wg ( B → A ) done by the gravitational force on the tomato as it travels from point B to point A is: Wg ( B → A ) = mgd cos 0° = mgd (76)
 B . Work done by a force in Lifting an object: (77) Consider an object of mass m that is lifted by a force F form point A to point B. The object starts from rest at A and arrives A m at B with zero speed. The force F is not necessarily constant during the trip. The workkinetic energy theorem states that: ∆K = K f − K i = Wnet We also have that K i = K f → ∆K = 0 → Wnet = 0 There are two forces acting on the object: The gravitational force Fg and the applied force F that lifts the object. Wnet = Wa ( A → B ) + Wg ( A → B ) = 0 → Wa ( A → B ) = −Wg ( A → B ) Wg ( A → B ) = mgd cos180° = mgd → Wa ( A → B ) = mgd Work done by a force in Lowering an object: In this case the object moves from B to A Wg ( B → A ) = mgd cos 0° = mgd Wa ( B → A ) = −Wg ( B → A ) = − mgd
 Work done by a variable force F ( x) acting along the xaxis: A force F that is not constant but instead varies as function of x is shown in fig.a. We wish to calculate the work W that F does on an object it moves from position xi to position x f . We partition the interval ( xi , x f ) into N "elements" of length ∆x each as is shown in fig.b. The work done by F in the j  th interval is: ∆W j = F j ,avg ∆x Where Fj ,avg is the average value of F N over the j th element. W = ∑ Fj ,avg ∆x We then take the limit of j =1 the sum as ∆x → 0 , (or equivalently N → ∞) N xf W = lim ∑ Fj ,avg ∆x = ∫ F ( x)dx Geometrically, W is the area j =1 xi between F ( x) curve and the x axis, between xi and x f xf (shaded blue in fig.d) W= ∫ F ( x)dx xi (78)
 The Spring Force: Fig.a shows a spring in its relaxed state. In fig.b we pull one end of the spring and stretch it by an amount d . The spring resits by exerting a force F on our hand in the opposite direction. In fig.c we push one end of the spring and compress it by an amount d . Again the spring resists by exerting a force F on our hand in the opposite direction The force F exerted by the spring on whatever agent (in the picture our hand) is trying to change its natural length either by extending or by compressing it is given by the equation: F = − kx Here x is the amount by which the spring has been extended or compressed. This equation is known as "Hookes law" k is known as "spring constant" F = − kx (79)
 x Work Done by a Spring Force O (a) Consider the relaxed spring of spring constant k shown in (a) By applying an external force we change the spring's x length from x i (see b) to x f (see c). We will O (b) calculate the work Ws done by the spring on the external agent x (in this case our hand) that changed the spring length. We O (c) assume that the spring is massless and that it obeys Hooke's law xf xf xf We will use the expression: Ws = ∫ F ( x)dx = ∫ −kxdx = −k ∫ xdx xi xi xi x 2 x2 kxi2 kx f f x Ws = −k = − Quite often we start we a relaxed 2 xi 2 2 spring (xi = 0) and we either stretch or compress the spring by an x kx 2 amount x ( x f = ± x). In this case Ws = − 2 (710)
 Three dimensional Analysis: r In the general case the force F acts in three dimensional space and moves an object on a three dimensional path from an initial point A to a final point B r The force has the form: F = Fx ( x, y, z ) i + Fy ( x, y, z ) ˆ + Fz ( x, y, z ) k ˆ j ˆ Po int s A and B have coordinates ( xi , yi , zi ) and ( x f , y f , z f ) , respectively r r dW = F ⋅ dr = Fx dx + Fy dy + Fz dz B xf yf zf W = ∫ dW = ∫ F dx + ∫ F dy + ∫ F dz x y z A xi yi zi z xf yf zf B W= ∫ F dx + ∫ F dy + ∫ F dz x y z O xi yi zi x path y A (711)
 WorkKinetic Energy Theorem with a Variable Force: Conside a variable force F(x) which moves an object of mass m from point A( x = xi ) dv to point B( x = x f ). We apply Newton's second law: F = ma = m We then dt dv multiply both sides of the last equation with dx and get: Fdx = m dx dt xf xf dv We integrate both sides over dx from xi to x f : ∫ Fdx = ∫ m dx xi xi dt dv dv dx dv dv dx = → dx = dx = vdv Thus the integral becomes: dt dx dt dt dx dt xf 2 m 2 x f mv f mvi2 W = m ∫ vdv = v = xi − = K f − K i = ∆K xi 2 2 2 Note: The workkinetic energy theorem has exactly the same form as in the case when F is constant! m F(x) W = K f − K i = ∆K O . A dx . B xaxis x (712)
 Power We define "power" P as the rate at which work is done by a force F . If F does work W in a time interval ∆t then we define as the average power as: W Pavg = ∆t dW The instantaneous power is defined as: P= dt Unit of P : The SI unit of power is the watt. It is defined as the power of an engine that does work W = 1 J in a time t = 1 second A commonly used nonSI power unit is the horsepower (hp) defined as: 1 hp = 746 W The kilowatthour The kilowatthour (kWh) is a unit of work. It is defined as the work performed by an engine of power P = 1000 W in a time t = 1 hour W = Pt = 1000 × 3600 = 3.60 × 106 J The kWh is used by electrical utility companies (check your latest electric bill) (713)
 Consider a force F acting on a particle at an angle φ to the motion. The rate dW F cos φ dx dx at which F does work is given by: P = = = F cos φ = Fv cos φ dt dt dt r r P = Fv cos φ = F ⋅ v v (714)
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