Work and kinetic energy. If a net force is applied on an object, the object may

Transcription

1 Work and kinetic energy If a net force is applied on an object, the object may CHAPTER 6 WORK AND ENERGY experience a change in position, i.e., a displacement. When a net force is applied over a distance, work is done. Work and kinetic energy work-kinetic energy theorem Work done by a variable force the dot product Power Potential energy conservative forces non-conservative forces conservative forces and the potentialenergy function equilibrium The work done is W = F x Δx, where the force is in the direction of the displacement, Δx = x x. In the lower figure, F x = Fcosθ. W = (ma x )Δx, But v 2 2 = v + 2ax Δx a x Δx = 1 2 (v2 2 v ). W = 1 2 mv2 1 2 mv 2. F x kinetic energy (K) This is known as the work-kinetic energy theorem W = K f K i = ΔK.

2 Dimension of work: F [M][L] [T] 2 and Δx [L]. Units: W [M][L]2 [T] 2 N m and/or Joule (J). (scalar). Question 6.1: A constant force of 1 N acts on a box of mass 2. kg for 3. s. If the box was initially at rest and the coefficient of kinetic friction between the box and the surface is µ k =.3, NOTE: Kinetic energy and work have the same dimension and units. (a) what is its speed of the box after 3. s? (b) How far did the box travel? Work can be positive or negative. If a force increases the speed of an object, the work done by the force is positive. If a force decreases the speed of an object (e.g., frictional force) the work done by the force is negative. (c) What was the work done by the applied force? (d) How much work was done by the frictional force?

3 N F = 1N m = 2. kg f k F Δx v = t = 3. s v =? mg (a) The net force acting on the box is F net = F f k = F µ k N = F µ k mg = 1 N (.3 2. kg 9.81 m/s 2 ) = 4.11 N. Therefore, the acceleration of the box is a = F net m = (4.11 N) (2. kg) = 2.6 m/s2. The final velocity after 3. s is given by v = v + at = 2.6 m/s 2 3.s = 6.18 m/s. (b) To find the distance traveled we have: v 2 = v 2 + 2aΔx, i.e., Δx = v2 v 2 = ( 6.18 m/s )2 2a 2 = 9.27 m m/s (c) The work done by the applied force is W = F.Δx = (1 N)(9.27 m) = 92.7 N m. (d) The work done by the frictional force is W f = µ k mg.δx = (.3)(2. kg)(9.81 m/s 2 )(9.27 m) = 54.6 J (or N m). Note the net work done by the applied force and the frictional force is W net = ( ) = 38.1 J. It is the net work that produces the increase in the speed of the box as we can see from the work-kinetic energy theorem. W net = ΔK = 1 2 mv2 1 2 mv 2 = 1 2 (2. kg)(6.18 m/s)2 = 38.2 J. (Rounding error.)

4 Let the mass of the motorcycle and rider be M. The work done is given by the work-kinetic energy theorem, i.e., W = ΔK = 1 2 M v f 2 2 v i. Question 6.2: Which of the following requires the most work to be done by the engine of a motor cycle? A: Accelerating from 6 km/h to 8 km/h. B: Accelerating from 4 km/h to 6 km/h. C: Accelerating from 2 km/h to 4 km/h. D: Accelerating from to 2 km/h. E: The same in each case. For A: ΔK = 1 2 M ( ) =14M. For B: ΔK = 1 2 M ( ) =1M. For C: ΔK = 1 2 M ( ) = 6M. For D: ΔK = 1 2 M ( ) 22 = 2M. So case A requires most work by the engine. Even though the change in speed is the same in each case, the work done is greatest for A because the initial speed is the greatest also.

5 F y F y = 8 N 3 m 6. kg mg (a) The work done by the applied force F y is F y Δy = (8 N)(3. m) = 24 J. (b) The work done by the gravitational force is Question 6.3: A box, of mass 6. kg, is raised a distance of 3. m from rest by a vertical force of 8. N. Find (a) the work done by the force, (b) the work done by the gravitational force, and (c) the final velocity of the box. ( mg)δy = (6. kg)(9.81 m/s 2 )(3. m) = 177 J. The work is negative because the gravitational force is in the opposite direction to the displacement. (c) The net work done on the box is W net = (F y mg)δy = 24 J 177 J = 63 J. Using the work-kinetic energy theorem, W net = ΔK = 1 2 mv2 1 2 mv 2 = 63 J, But v =. v 2 = 2(63 J) 6. kg = 21. (m/s)2, i.e., v = 4.58 m/s.

6 Question 6.4: A car accelerates from rest and gains a certain amount of kinetic energy. The Earth A: gains more kinetic energy. B: gains the same amount of kinetic energy. C: gains less kinetic energy. D: loses kinetic energy as the car gains it. The work-kinetic energy theorem tells us that the kinetic energy gained is equal to the work done, i.e., the product of the force and the displacement. By Newton s 3 rd Law, the force exerted by the car on the Earth is equal and opposite to the force exerted by the Earth on the car. However, the distance over which these forces act are not equal; the Earth moves a negligible amount as a result of the cars motion. Therefore, the answer is C: the Earth gains less kinetic energy.

7 If F is not parallel to the displacement, we have to use the component of the force parallel to the displacement: θ F i.e., the work done W = ( F cosθ) Δ s = F Δ s cosθ. θ B A Δ s The dot or scalar product of two vectors, A and B is defined as A B = A B cosθ. If A (A x, A y,a z ) and B (B x,b y,b z ) Then A B = A x B x + A y B y + A z B z ~ see revision notes on website ~ Question 6.5: A force object and produces a displacement Find F = (2î 2 ĵ+ 2ˆk) N (a) the work done by the force, and (b) the angle between F and Δ s. acts on an Δ s = (2î + ĵ 3ˆk) m. Therefore, the work done is W = F Δ s

8 (a) The work done by the force is W = FiΔ s = F x Δs x + F y Δs y + F z Δs z What is the significance of the negative sign? (b) Also, = (2 N)(2 m) + ( 2 N)(1 m) + (2 N)( 3 m) = 4 J. FiΔ s = F Δ s cosθ, i.e., cosθ = FiΔ s F Δ s, where θ is the angle between F and Δ s. Now F = (2 N) 2 + ( 2 N) 2 + (2 N) 2 = 3.46 N. Δ s = (2 m) 2 + (1 m) 2 + ( 3 m) 2 = 3.74 m. θ = cos 1 FiΔ s F Δ s = cos 1 4 J (3.46 N)(3.74 m) W = F x Δx shaded area. In the preceding, the applied force has been constant. How do we handle the situation if the force varies? Work done over the small displacement Δx i is F xi Δx i. Since = < θ < 27, then W <. W x1 x 2 = Limit Δx i x ( i F xi Δx i ) 2 = F xdx, i.e., the area under the curve between x 1 and x 2. x 1

9 We find the work done over the displacement and then use the work-kinetic energy theorem to determine the speed. Question 6.6: The force F x shown above acts on an object of mass 2 kg. If the object is at rest at x =, what is its speed at x = 1 m? The work done is the area under the plot, i.e., [1] + [2] + [3] = ( ) J = 46 J. Therefore, using the work-kinetic energy theorem But v =. v = ΔK = 1 2 m(v2 v 2 ) = 46 J. 2ΔK m = 2(46 J) 2 kg = 6.78 m/s.

10 Question 6.7: A force, F = 2x 2 î + 3yĵ 2zˆk ( ) N, where x, y and z are in meters, acts on an object of mass 5. kg. (a) How much work is done on the object if it moves from point A, (1 m,2 m,1m), to point B, (2 m,3 m,2m)? (b) If the speed of the object at point A is 1. m/s, what is it speed at point B? (a) The work done from A to B is B A B A ( ) W AB = Fid s = F x î + F y ĵ+ F z ˆk i dxî + dyĵ+ dzˆk x B y B z B = F x dx + F y dy + F z dz. x A y A z A W AB = 2x 2 dx + 3ydy 2zdy 2 = 2 x y2 2 1 (b) Using the work-kinetic energy theorem: 2 1 ( ) W AB = ΔK = K B K A, i.e., K B = K A + W AB z2 2 = = 9.2 J. 1 2 K B = 1 2 (5. kg)(1. m/s) J = 11.7 J = 1 2 (5. kg)v B 2. v B = 2(11.7 J) 5. kg = 2.16 m/s.

11 POWER Power is the rate at which work is done or energy is dissipated (i.e., quickly or slowly). For example, running up a flight of stairs quickly requires more power than walking up the same flight of stairs slowly. At any instant, the instantaneous power is P = dw F d s = = F d s dt dt dt = F v (scalar), so power can be > or < depending on the angle between F and v. Dimensions: Power work done time = [M][L]2 [T] 3. [M][L]2 [T] 2 [T] Question 6.8: (Revisiting question 5.5.) The driver of a 12 kg car moving at 15. m/s is forced to slam on the brakes. The car skids to a halt after traveling a distance of 25.5 m. We found the coefficient of kinetic friction was µ k =.45 and it took 3.4 s for the car to stop. (a) How much work is done by the frictional force? (b) What is the average power dissipated by the frictional force? Units: J/s watts (W) 1 W 1 kw 1 HP 746 W

12 v " v = f k N (a) Two ways to find the work done by the frictional force: x " x (i) W f = f k.l = (µ k mg)l =.45(12 kg)(9.81 m/s 2 )(25.5 m) = J. (ii) W f = ΔK = 1 2 mv2 1 2 mv " 2 = 1 2 (12 kg)(15. m/s)2 = J. (b) Average power dissipated by the frictional force is P av = W f t l = J 3.4 s = W = 39.7 kw. The power is < as W f <, which means that energy is removed from the system (as the car is slowing down). mg One-dimensional example: P = F v = F x v x = ma x v x i.e., a x = P mv x. So, if the engine operates with constant power output, the resulting acceleration is inversely proportional to the velocity, i.e., as v increases, a decreases. Also P = F x v x = ma x v x = mv x dv x dt = d 1 dt 2 mv x 2 = dk dt, i.e., the instantaneous power at some time t is the rate of change of kinetic energy at that same time.

13 (a) Power is related to the change in kinetic energy ΔK over a time interval Δt, viz: So, at constant power, Also, since v =, ΔK = 1 where M is the mass of the 2 Mv2, car, and v is the final speed. P = ΔK Δt. Δt ΔK. Question 6.9: A car, with mass 1 kg, accelerates from zero to 2 mi/h in 2 s. Assuming the power of the engine is constant (a) how long would it take to accelerate from zero to 6 mi/h? (b) What is the power generated by the engine? From to 2 mi/h: ΔK 1 = 1 2 M(2)2 = 2M. From to 6 mi/h: ΔK 2 = 1 2 M(6)2 = 18M. ΔK 2 = 9ΔK 1, i.e., Δt 2 = 9Δt 1. Since Δt 1 = 2 s then Δt 2 = 18 s. So, the time required to increase the speed by a factor of n, increases by n 2. (b) Convert 2 mi/h to m/s: (2 mi/h)( m/mi) 36 s/h = 8.94 m/s.

14 Therefore, the power generated by the engine is Check: P = ΔK 1 = 1 (1 kg)(8.94 m/s) 2 Δt s = W 26.8 HP. 6 mi/h (6 mi/h)( m/mi) 36 s/h = 26.8 m/s. ΔK 2 = 1 2 (1 kg)(26.8 m/s)2 = J. Question 6.1: A force, F = (5.4î ĵ) N, acts on a 3. kg object. If the object is initially at rest, what is the instantaneous power delivered by the force after 2 s? P = ΔK 2 = J Δt 2 18 s = W 26.8 HP.

15 The acceleration of the object is a = F (5.4î ĵ) N = = (1.8î m 3. kg +.9 ĵ) m/s2. The velocity after 2 s is v = v " + at = (1.8î +.9 ĵ)(2 s) = (3.6î +1.8 ĵ) m/s, so the instantaneous power is P = Fi v = (5.4î ĵ)i(3.6î +1.8 ĵ) W = 24.3 W. Question 6.11: A 5. kg object starts from rest at x = and moves along the x-axis under the influence of a single force where F x F x = x 3 2.x , is in Newton s and x is in meters. (a) What is the work done as the object moves from x = to x = 3 m? (b) Find the power delivered by the force as the object passes the point x = 3 m?

16 (a) The work done is = x4 3 W = (x 3 2.x )dx 4 x x 3 (b) The instantaneous power is P = Fi v F x v x. We find v x using the work-kinetic energy theorem: W = ΔK = K f K i = 1 2 mv x 2. v x = 2W m = 2(11.3 J) 5. kg 3 = 11.3 J. = 2.13 m/s. F x = (3. m) 3 2.(3. m) = 12. N. P = (12. N)(2.13 m/s) = 25.6 W. Potential energy Consider lifting an object through a distance h; how much work is done? The net force is F m g = m a y. The work done by you in lifting the object through a distance dy is dw = Fid y = Fdy, so the total work done is W = Fdy = (mg + m a y )dy = mg dy + m a y dy. But h h h i.e., the object starts and finishes at rest. So the work done by you against the gravitational force depends only on h and not on the rate the work was done. h h v m a y dy = m dv dy = m dy dv = m v dv dt dt h = 1 2 mv2 v v v v v =, if v = v ( = ), h W = mg ydy = mgh.

17 What about lifting the object at an angle θ to the horizontal? The net force along the sloped line (s) is F mgsinθ = ma s, where is the acceleration along s and mgsinθ is the component of the weight force. But, if the object starts and finishes at rest, the second integral is zero, as we showed on the previous slide. Also, since h = lsinθ then dh = sinθdl, and so the work you do is i.e., the same as before, so the work done is independent of the route l F h mg θ l W = (mgsinθ + ma s )dl = mg sinθdl + m a s dl. h l W = mg dh = mgh, a s l Although we have done work on the object in both cases there is no change in kinetic energy. So, what s happened to the work? Clearly the configuration has changed since the object is in a different position after the work was done. In fact, the gravitational potential energy of the object has changed. The amount of potential energy gained (ΔU G ) is equal to the work we do, i.e., ΔU G = (U 2 U 1 ) = mgh. When the object is raised, the work done by the gravitational force is W 12 = ( mg)h = ΔU G = (U 2 U 1 ). [2] F In the raised position, the object is capable of doing work; for example, when released and allowed to fall, its kinetic energy increases so it could strike a nail and drive it into the floor The work done by the gravitational force when the object falls to its original position is W 21 = ( mg)( h) = mgh. mg [1]

18 ΔK = W 21 = mgh So the net work done by the gravitational force is zero, i.e., The gain in kinetic energy of the object as it falls is i.e., W 21 = W 12. ΔK = W 21 = mgh, 1 2 mv2 = mgh v = 2gh. A similar situation arises when you compress a spring, e.g., in a toy gun. The spring is compressed by applying two equal and opposite forces. The net force on the spring is zero so there s no change in kinetic energy, but you have done work on the spring. What s happened to the work? Clearly, the length of the spring has changed. In fact, the spring has stored the work you did as elastic potential energy. The increase in potential energy is ΔU E = 1 2 kx2, where k is the spring constant a measure of the stiffness and x is the change in length of the spring. When released the spring will transfer the potential energy gained to the kinetic energy of the ball. The net work done by the elastic force in this example is zero.

19 Conservative forces What about non-conservative forces? If the work done by a force in going from configuration 1 to configuration 2 is and the work done in going W 12 from configuration 2 to configuration1 is W21, then, if W 21 = W 12 the force is defined as a conservative force. The gravitational and the elastic forces are examples of conservative forces. Complementary definition: The work done on an object by a conservative force is zero when the object returns to its initial position, i.e., when the object moves around any closed path, no matter the route. Consider pushing a book across a table from position A to position B. You have to do work against the kinetic frictional force between the book and table. Then W AB = F k d A B = mgµ k d A B. Consider two different paths, a and b. Since b > a, then W b > W a. So, the work done against the frictional force does depend on the path. Therefore, the frictional force is a non-conservative force. NOTE: the work done by a non-conservative force is non-recoverable (producing heat, sound, etc.).

20 Question 6.12: In a region of space, the force on an electron is given by Where c is a constant. The electron moves round a square loop in the x-y plane. If the corners of the loop are at how much work done on the electron by the force while completing one trip around the loop? Is the force a conservative force? F = cyî (,), (l,), (l,l), (,l), (,l) dlĵ y F = clî F dlî dlĵ (l,l) dlî (,) F = (l,) From (,) to (l,) the force F =. W 12 =. From (l,) to (l,l) the force increases linearly but F d l, so W 23 =. Similarly, W 41 =. From (l,l) to (,l) W 34 = Fid l = clîidxî = cl dx = cl 2. So, the total work done is l l W = W 12 + W 23 + W 34 + W 41 = cl 2. l x F Since the start and end points are the same, the work done by the force around the loop is non-zero. Therefore, F is a non-conservative force. Note, if c < the electron gains energy as it goes around the loop.

21 Potential energy function A property of a conservative force is that there is a potential energy function U(r) associated with the force, which tells us how potential energy varies with position. As we ve seen with the gravitational force, the work done can be expressed as the difference between the initial and final values of the potential energy, i.e., W A B = ΔU = (U B U A ). Writing this equation in incremental form dw = du = Fid s, where U is the potential energy function. Therefore, the change in potential energy from point A to point B is ΔU = U B U A = du = Fid s. Since du = Fid s, then in one-dimension, B A B A du = F x dx i.e., F x = du dx. What this means is, if we know the functional form of U(x), we can determine the force at any point. Conversely, if we know the functional form of the force F x, we can find the potential energy function since As an example, the potential energy function for a compressed spring is Therefore, the force associated with this potential energy function is which is Hooke s Law, where x is the compressed distance. When x =, F =, du the spring is relaxed and This dx =. is known as equilibrium (chapter 4). At equilibrium F x du = F x dx i.e., U = F x dx. U(x) = 1 2 kx2. F(x) = du(x) dx du dx =, = kx, so U(x) is an extremum.

22 du However, does not uniquely define a minimum in dx = the potential energy function. STABLE EQUILIBRIUM Equilibrium position x. U(x ) is a minimum. UNSTABLE EQUILIBRIUM Equilibrium position x. U(x ) is a maximum. d 2 U dx 2 > d 2 U dx 2 < Question 6.13: A potential energy function varies as U(x) = 2x 4 x 2, where U is in Joules and x in meters. (a) At which point(s) is this object in equilibrium? (b) What is the state of equilibrium at these positions? NEUTRAL EQUILIBRIUM Equilibrium position x. U(x ) is a an inflexion point. d 2 U dx 2 =

23 (a) Equilibrium occurs where F = du dx =. du dx ( = 8x3 2x), therefore, the equilibrium points are solutions of the U(x) = 2x 4 2x equation: ( 8x 3 2x) = 2x 4x2 1 ( ) =. The solutions are x = and x = ±.5 m. (b) For the state of equilibrium, find the sign of d 2 U dx 2 = 24x2 2. When x =.5 m, d 2 U = 6 2 = 4, i.e., > (stable). 2 dx Stable Unstable Stable When x =, d 2 U = 2, i.e., < (unstable). 2 dx When x = +.5 m, d 2 U = 6 2 = 4, i.e., > (stable). 2 dx