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Lecture6 - Work Work, Energy and Power Chapter...

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Unformatted text preview: Work Work, Energy and Power Chapter 7 & 8 Imagine the following situa>ons: 1.  2.  3.  You are liCing a 20 lb box onto a shelf 3C off the ground. You are liCing a 20 lb box onto a shelf 6C off the ground. You are liCing a 10 lb box onto a shelf 3C off the ground. Which one requires more force ? What about the height you have to liC it – does that make any difference ? Need to define new quan>ty which includes force and distance over which force is applied it is WORK. Work done by a constant force Work done by a force is the scalar product of force and the displacement: W = F ⋅ Δr = F Δr cosθ θ Δr The SI unit of work is joule (J), which is the product of the units of force and displacement, i.e., 1 J = 1 Nm. Note: •  Work is a scalar. •  If the angle between force and displacement is 90°, no work is done (an example?) •  If θ = 180°, W<0. this simply means work is done against the force in ques>on by some other agent (force). (an example?) Example You are pushing an object with a force that is directed 30o downward from the horizontal and has a magnitude of 56 N. How much work do you do aAer pushing the object for 10m horizontally ? W == F ⋅ Δr = F Δr cos(−30 o ) = 485 Nm 300 Example You are playing tug of war with a friend. You apply a force of 60 N, while your friend applies a force of 80 N in the opposite direcFon and pulls you 3 m in her direcFon. How much work do you do ? 60 N 3 m Once again, work can be nega>ve. Here, in this example, work done by you is nega>ve. Work done by your friend is posi>ve (W = (80N)(3m) = +240 J. We need to be clear who or which force is doing the work. Example You are accelera>ng an object of mass m from rest to a speed of v m/s. How much work have you done on the object ? To accelerate an object you need to apply a force: Assume uniform accelera>on: v 2 − vi2 v2 Δr = Δx = x − xi = = 2a 2( F / m ) ⇒ W = F Δx = 12 mv 2 Work needed to accelerate an object depends on the change in velocity squared. We will learn soon that this is expression for kine>c energy gained by the object. Example During a storm, a crate is sliding across a slick, oily parking lot through a displacement of d = ( ­3.0 m)i while a steady wind pushes against the crate with a force of F = (2.0 N)i+( ­6.0N)j. How much work does the wind do on the crate ? W = F ⋅ d = (−3.0 m ) ⋅ (2.0 N ) + (0 m )(6.0 N ) = −6 J Example Work done by gravity A liele girl slides down her backyard slide an angle θ w.r.t ver>cal. How much work does the gravita>onal force do on her? W = Fg ⋅ Δr = Fg Δr cos(θ ) = mg Δr cos(θ ) = mg Δy i.e., W = m gh , Fg Δy θ Δr where h is the height of the slide. Note: •  Work done by gravity does not depend on how long is the slide, on whether it is curved or looped, as long as the the height is the same. •  If a person or object is moving up, gravity does nega=ve work, which means work is done against gravity. Work done by a varying force W =F · Δr (= Fx Δx + Fy Δy + Fz Δz in 3D space) assumes that F is constant. This is not always the case. How would we calculate the work if the force would change with distance ? F(x) For a small Δx we can approximate the work done by FΔx, where F is the average force during the displacement Δx. Then to find the sum of all such FΔx’s. The smaller Δx is, the better will be the approximation. In the limit Δx 0, this sum becomes the integral of F(x) from xi to xf: xf W = F ( x f − xi ) xi F Δx xf W = F Δx x W= xi ∫ F ( x ) dx This formula can be generalized as: W = ∫ F i dr The integral is evaluated over the path that takes the particle through the space. This integral is an example of what is known as a line integral. Work done by a spring The spring force: If you stretch or compress a spring from its equilibrium state, there will be a force ac>ng in a direc>on to restore the equilibrium state of the spring. Hooke’s law: F = − k x Here x is the displacement away from the equilibrium (relaxed) state of the spring. k is the measure of the ‘strength’ of the spring, it is called the spring constant. (More ridgid springs have larger ‘k’ values) The minus sign indicates that the force is in the opposite direc>on of the displacement, i.e. the force acts to restore the spring to its relaxed state. Now since we know how force varies with x, we can find the work done by the spring: W= −x ∫ F dx′ = 0 −x ∫ (− kx′) dx′ → W = 0 12 kx 2 For any arbitrary displacement, the work done by the spring is: W= 1212 kxi − kx f 2 2 And the work done on the spring by the applied external force is: Wapp = 1212 kx f − kxi 2 2 Kine>c energy, work ­kine>c energy theorem Consider a system consis>ng of a single object, like a block shown in the figure. Suppose the net force ac>ng on it is Fnet ac>ng along the x ­axis, and the infinitesimal displacement is dr = dx î . The net work done on the block by this force is: Wnet dv dv dx = ∫ F dx = ∫ ma dx = m ∫ dx = m ∫ dx = m ∫ v dv dt dx dt xi xi xi xi vi 1212 mv f − mvi .......................(1) 2 2 2 xf xf xf xf vf → Wnet = 1 m The quan>ty v is an important quan>ty. It is called the kineFc energy of the par>cle 2 and represents the energy associated with its mo>on. It is oCen denoted by KE or simply by K: 12 K= Equa>on (1) says that the work done by a net force on a par>cle of mass m is equal to the change in the par>cle’s kine>c energy. Wnet = K f − K i = ΔK 2 mv This last statement is known as the work ­kine:c energy theorem. We may state is as: When work is done on a system and the only change in the system is its speed, the net work done on the system equals the change in kineFc energy of the system. Units of energy The SI unit of energy is joule (J). It is clear from work ­kine>c energy theorem that kine>c energy (hence energy in general) an work must have the same units. Example: Work done in liCing an object. When we liC an object, we have to apply a force, FA. At the same >me the gravita>onal force Fg is ac>ng on the object as well. ΔK = K f − K i = W = ( FA + Fg )id = WA + Wg Assume that ini>ally the object is at rest and also that the object is at rest again at the end of the liC. ΔK = WA + Wg = 0 ⇒ WA = −Wg = −( mgd cos180°) ⇒ WA = mgh The work done to liC an object by a height h is given by : W = mgh Example 7.6 (textbook): A 6.0 ­kg block ini>ally at rest is pulled to the right along a horizontal, fric>onless surface by a constant horizontal force of 12 N. Find the speed of the block aCer it has moved 3.0 m. The net external force ac>ng on the block is the horizontal force F = 12N. Work done by this force is W = F Δx = (12 N )(3m) = 36 J Now, using work ­kine>c energy theorem, we get W = ΔK = K f − K i = 1 ⇒ mv 2f = W 2 ⇒ vf = 2W ( 2)(36 J ) = = 3.5m / s m 6.0 kg 121212 mv f − mv i = mv f − 0 2 2 2 Example A 20 kg box is pulled up a fric>onless ramp over a distance of d=7m. If the ramp makes an angle of 300 with the horizontal, how much work was done ? Method 1: If the box is moving at constant velocity, tension in rope would be T = mgsinθ Then the work would be W = F i d = T i d = mg sin(θ ) d d =7m m mg Method 2: We can directly use the the ver>cal height it is raised to. h = d sin q = (7 m)(sin 300) = 3.5 m W = mgh = 20 ⋅ 9.8 ⋅ 3.5 = 686 J T θ h → W = 20 ⋅ 9.8 ⋅ 7 ⋅ sin(30 ) = 686 J Example: Centripetal mo>on Puck on string, moving in circle at constant speed. What is the work done by the tension in the string? v2 T = − m (cosθ iˆ + sin θ ˆ) j R points to the center ˆ dr = v dt = v dt (− sin θ i + cosθ ˆ ) j Points along the tangent v3 W = ∫ T ⋅ dr = − m ∫ ( − sin θ cosθ + sin θ cosθ ) dt = 0 R No work is done! Example A par>cle experiences a force changing only the kine>c energy of the par>cle. The par>cle moves from posi>on (2m, 3m) to (3m, 0). What is the work done on the par>cle, and does its speed increase ? W = ∫ F i dr = ∫ ( Fx dx + Fy dy ) = 3 0 2 xf xi ∫ F dx + ∫ F dy x y yi yf = ∫ 3x dx + ∫ 4 dy = ⎡ x 3 ⎤ 3 + ⎡ 4 y ⎤ 0 ⎣ ⎦2 ⎣ ⎦3 2 3 = ⎡33 − 23 ⎤ + ⎡ 4 ⋅ 0 − 4 ⋅ 3⎤ = 7.0J ⎦ ⎣ ⎦⎣ Since W is posi>ve, energy is transferred to the par>cle. Thus its kine>c energy is increased (by 7 J) and its speed must have increased also. Poten>al energy A system is said to have energy it is capable of doing work. A moving object can do work. The energy it possesses just because of its mo>on is its kine>c energy. It is capable of doing work, for example it can hit an object and deform it or make it move or does both. We already know that when work is done on a body, the kine>c energy is added to the body equal to the work done. Imagine a stone held in your hand. When you release the stone, it does the work (strictly, gravita>onal force does the work here). The stone has energy because of its posi>on in presence of gravita>on force. Consider also, a loaded spring. When released, it does work and generates mo>on in the body aeached with it. In this example, the body aeached at the spring (or the spring itself) has energy because of the special configura>on of the spring. In the last two examples, we note that because of the posi>on of the object and the forces that are present, there is a type of “stored” or potenFal energy present in the system. This energy can be released (or “unstored”) and turned into mo>on, a.k.a. kine>c energy. The released stone, gains kine>c energy at the expense of its poten>al energy. Ul>mately, before hipng the floor all the poten>al energy is converted to kine>c energy. Poten>al Energy & Work You are pushing on a block aeached to a spring, compressing the spring. The work done by spring force is: W = − k x 2 As we compress the spring however, the poten>al energy, U, of the spring has increased. Therefore: ΔU = −W 1 2 Take another example. You raise a body of mass m through a height h. Work done by gravita>onal force is W = − mgh . You have increased the poten>al energy of the body: ΔU = −W Poten>al energy stored in a spring In our previous example of compressed spring, if we take U = 0, at x = 0, then ΔU = U ­0 or U = ΔU. Since ΔU =  ­W, the elas>c poten=al energy stored in a spring is given by Us = 12 kx 2 Gravita>onal poten>al energy In the same way, if we assume that poten>al energy at the floor is zero in our example of a body raised to a height h, we get the following expression for the gravita>onal poten>al energy of the body: U g = mgh Conserva>ve forces Take a body of mass m. Raise it from the floor to the height h. The work done by gravita>onal force is –mgh. Then release it. The work done by gravity is mgh. In the overall up and down mo>on work done by gravity is: W = − mgh + mgh = 0 In this example, by reversing the path the work is “recovered”. Because of this, the gravita>onal force is a conservaFve force. The fact that gravita>onal force is conserva>ve also means that the work done by gravity in moving from point A to point B does not depend on the path taken. It only depends on the posi>on A and B. Spring force is also a conserva>ve force. Work done by spring only depends on the ini>al and final posi>ons. Work done by a conserva>ve force (eg, gravita>onal force, spring force, etc) on a par>cle moving through any closed path is zero. Non ­conserva>ve forces (a) fk (b) fk Sliding a block against fric>on. How much work does the fric>on do? Wa = − fk ( x2 − x1 ) < 0 Wb = fk ( x1 − x2 ) < 0 → Wa + Wb ≠ 0(< 0 ) i.e. the fric>onal force always does nega>ve work, slowing the block down. We cannot recover the work by simply going the other way! Fric>on is a non ­conservaFve force. Conserva>on of mechanical energy The term mechanical energy means the sum of kine>c and poten>al energy of a system. What happens to the mechanical energy, when we do work on the system? (Assume conserva>ve forces) We know, Therefore, ΔE = ΔK + ΔU = W − W = 0 The mechanical energy in an isolated conserva>ve system is conserved. When work is done, it converts poten>al energy and kine>c energy into each other, but the sum of the two is unchanged. Example A body drops from a height y = h. Calculate the total mechanical energy at any point on its path. Refer to the figure. y At the top point A, K = 0 U = mgH → E = mgH 1 1 K = mv 2 = m(0 + 2 gH ) = mgH 2 2 U=0 E = mgH 121 mv = m [ 0 + 2 g( H − y)] = mgH − mgy 2 2 U = mgy E = mgH K= K = 0 E = U H ­y E = K+U A At the boeom B, P y=H y U = 0 B E = K y = 0 At any point P, Note that P is an arbitrary point between A and B (included). Clearly, poten>al energy changes to kine>c energy, but the total energy remains the same. If we throw the ball upward and watch the upward mo>on, kine>c energy changes into poten>al energy. Example: Chapter 8 Problem 4 Chalk board stuffs Example Conversion between kine>c and poten>al energy is more interes>ng in simple pendulum (also in mass ­spring system). This can more clearly be represented in energy diagram. x Poten>al energy curve What if the total energy is less than the maximum of the poten>al ? As the par>cle approaches a ‘turning point’ (intercept of E and U), its kine>c energy K = E ­U goes to zero and it cannot go beyond that point and it turns back as it seeks the minimum poten>al energy configura>on. If there is a turning point on the other side, the par>cle’s mo>on is confined between the turning points. This means that the par>cle undergoes oscillatory mo>on. U Turning points E x Equilibrium points Neutral equilibrium: force remains zero, kine>c energy remains zero Unstable: Force is non ­zero and points away from equilibrium point. Par>cle can gain kine>c energy, since E is larger than U at both sides of equilibrium point U(x) E E Stable equilibrium: Force is non ­zero, but points in direc>on to return par>cle to equilibrium point. Par>cle can not gain kine>c energy as poten>al energy curve is above total energy away from the equilibrium point. E x Poten>al energy and force If F is a conserva>ve force, we know that: How can we find the force if we know the poten>al energy ? Inverse of integra>on is differen>a>on F(x) = − dU ( x ) ...........(1) dx In other words, force is the nega>ve poten>al gradient. For example, for spring U= 12 dU k x → F(x) = − = − kx 2 dx The plot of U(x) vs x is oCen called the poten>al energy curve. From (1) we see that the slope of such a curve gives the force. Zero force means either maximum or minimum of the curve. The sign of the second deriva>ve determines whether a maximum or minimum of the curve. Let’s go back to the equilibrium: For equilibrium: dU F = − = 0 dx For stable equilibrium we must also have dF d 2U <0⇒ 2 <0 dx dx In stable equilibrium, the par>cle remains within the “poten>al well”. Poten>al energy curves play an important role in chemistry, biology, and materials science, to study bonding between atoms, molecules and nano ­par>cles. A ques>on… A person (700 N weight) is running up some stairs of height 5 m. IniFally, the person runs up the stairs in 10 seconds, the second Fme the person walks up the stairs, taking 5 minutes. In which case does the person do more work ? Answer: the work done in both cases is equal because the heights are the same If the person does the same amount of work, why does he feel more Fred when running up the stairs rather than walking up ? It depends on the rate at which he does the work. We need a quan>ty that gives the >me rate of work done. This quan>ty is power. Power Power is defined as the >me rate of energy transfer: P= dE dt In our current context of mechanical energy, power can also be defined as the >me rate of work done: dW P= dt Since dW = F.dr, for constant force F we can write d dr P = ( F idr ) = F i dt dt → P = F iv The SI unit of power is wae (W). 1 W = 1 J/s. A commonly used unit of power is horsepower. 1 horsepower = 746 W Our electricity provider uses uses kilowae ­hr as a unit of energy we use. 1 kw ­hr = (103 W)(3600 s) = 3.60×106J Systems: isolated and non ­isolated If energy is transferred across the boundary of a system, we say the system is non ­isolated. If energy is not transferred across the boundary of a system, we say the system is isolated. Energy is conserved within an isolated system. If work is done by conserva>ve forces within the system, the total mechanical energy is constant. In other words, in an isolated system the change in mechanical energy is always zero since no energy is gained or lost through transfer, as long as the forces involved are conserva>ve For a non ­isolated system an external force will do WORK ON the system: Posi=ve work increases energy of system. (“Work done on the system by surroundings”) Nega=ve work decreases energy of system. (“Work done by the system on surroundings” Work done by an external force Case 1: No fric>on (conserva>ve forces only) We already know that in absence of dissipa>ve forces like fric>on, we have Wext = ΔEmec = ΔK + ΔU Case 2: With fric>on (in presence of non ­conserva>ve forces) F − fk = ma ⇒ ∫ ( F − fk ) dx = xi xf xf xf xi ∫ ma dx vf dv ⇒ Wext − fk d = ∫ m dx = dt xi dx m dv = ∫ dt vi vf vi ∫ mv dv = 1212 mv f − mvi = ΔK 2 2 Here d = xf –xi. Thus we have ΔK = W − f d ext k This is the work ­kine>c energy theorem generalized to include fric>onal forces, which are non ­conserva>ve forces. Change in total energy of the system is then ...
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