Potential energy functions used in Chapter 7

Transcription

1 Potential energy functions used in Chapter 7 CHAPTER 7 CONSERVATION OF ENERGY Conservation of mechanical energy Conservation of total energy of a system Examples Origin of friction Gravitational potential energy The gravitational potential energy of a mass M at a distance h above the zero of potential energy (usually the ground is U G = Mgh. Elastic potential energy When a spring is compressed (or stretched) a distance l from its equilibrium (i.e. unstretched) length the elastic potential energy of the spring is U E = 1 2 kl2 where k is the spring constant.

2 Conservation of mechanical energy Consider the motion of an object in one dimension. In the previous chapter we found that the work done by a force moving the object from point 1 to point 2 is where K is the kinetic energy. We also found that if the force is conservative then W 12 = Fdx = ΔK = K 2 K 1 where U is the potential energy Fdx = ΔU = (U 2 U 1 ) 1 ΔK = K 2 K 1 = ΔU = (U 2 U 1 ) i.e. K 2 + U 2 = K 1 + U 1. Defining the total (mechanical) energy as E = K + U then ΔE = (K 2 K 1 ) + (U 2 U 1 ) = 0 i.e. the total energy remains unchanged. This is a statement of the conservation of mechanical energy. Conservation of mechanical energy provides us with an alternative way to determine velocities displacements etc. as illustrated in the following. An object is fired vertically from the ground with an initial velocity v!. What is its velocity at a height y above the ground? What is the maximum height h it reaches? Let the gravitational potential energy at ground level U 1 = 0. The kinetic energy h y of the object at ground level is K 1 = 1 2 mv! 2 so the mechanical energy is E 1 = K 1 + U 1 = 1 2 mv! 2. Since the gravitational force is conservative mechanical energy is conserved so we put E 1 = E.

3 At a distance y above the ground the kinetic energy is K 2 = 1 2 mv2 and the gravitational potential energy is U 2 = mgy. Since mechanical energy is conserved K 2 + U 2 = 1 2 mv2 + mgy = E = 1 2 mv! 2 y U G = mgy K = 1 2 mv2 i.e. v 2 = v! 2 2gy. 2 v = v! 2gy. At maximum height v = 0 so K 3 = 0. Also U 3 = mgh. Using the conservation of mechanical energy K 3 + U 3 = mgh = E = 1 2 mv! 2. h = v! 2 2g. See how the individual components of the mechanical energy vary in time as the object rises. Since E = K + U constant (in time) ΔE = ΔK + ΔU = 0. ΔK = ΔU i.e. the decrease in kinetic energy equals the increase in gravitational potential energy (and vice versa when an object is dropped). We would get the same results using the kinematic equations derived in chapter 2.

4 h Question 7.1: Two children throw stones from the top of a building at the same instant with the same initial speed. One is thrown horizontally the other is thrown vertically upward. Which of the following statements best describes what happens. The stones will strike the ground A: at the same time with equal speeds B: at different times with equal speeds C: at the same time with different speeds D: at different times with different speeds. This involves the conservation of mechanical energy. The change in gravitational potential energy (ΔU = mgh) is converted to kinetic energy. Therefore for each stone: Putting U f = 0 at ground level we get 1 2 mv f 2 = 1 2 mv i 2 + mgh so their final speeds are the same as they started from the same height! Since the speeds are the same at the ground we have where t 1 K f + U f = K i + U i. i.e. v f 2 = vi 2 + 2gh v f = v i 2 + 2gh 0 gt 1 = v y gt 2 i.e. t 1 = t 2 v y g is the time for the stone thrown horizontally and t 2 is the time for the stone thrown vertically. So t 1 < t 2. Therefore the answer is B different times with equal speeds (and independent of mass).

5 The following two examples involve two forms of potential energy; gravitational and elastic. Note that both are associated with conservative forces and so we can use the conservation of mechanical energy. Before [1] y 1 = 0.08 m v 1 = 0 m = 1.0 kg After [2] y 2 =? v 2 = 0 k = 1200 N/m We choose the ground to be the zero of gravitational Question 7.2: A 1.0 kg block is placed on top of a vertical spring with its lower end fixed to the ground. The spring has an unstretched length of 20.0 cm and spring constant k = 1200 N/m. The spring is then compressed to a length of 8.0 cm and released so the block is shot vertically upwards. What height did the block reach? potential energy. At position [1] K 1 = 0 : U G1 = mgy 1 : U E1 = 1 2 k(δy)2 and at position [2] K 2 = 0 : U G2 = mgy 2 : U E2 = 0. Using the conservation of mechanical energy K 2 + U G2 + U E2 = K 1 + U G1 + U E1 i.e. mgy 2 = mgy k(δy)2. y 2 = y k mg (Δy)2 (1200 N/m)(0.12 m)2 = (0.08 m) + 2(1.0 kg)(9.81 m/s 2 = 0.96 m. )

6 k = 400 N/m 5.00 m x Question 7.3: A 3.00 kg object is released from rest at a height of 5.00 m on a curved frictionless ramp as shown above. At the foot of the ramp is a spring with a spring constant k = 400 N/m. The block slides down the ramp compressing it distance x before coming momentarily to rest. What is the distance x? k = 400 N/m 5.00 m U G = 0 x At the top of the ramp: K = 0 : U G = mgh : U E = 0. E top = mgh = (3.00 kg)(9.81 m/s 2 )(5.00 m) = J. At the bottom of the ramp: K = 0 : U G = 0 : U E = 1 2 kx2. E bottom = 1 2 (400 N/m)x2 = 200x 2 J. Mechanical energy is conserved E bottom = E top i.e. 200x 2 = J. x = 0.86 m.

7 Question 7.4: An object of mass 2.5 kg is confined to move along the x-axis. If the potential function is U(x) = 3x 2 2x 3 where U(x) is in Joules and x is in meters and the velocity of the object is 2.0 m/s at x = 1 m what is its velocity at x = 1 m? Since a conservative force is involved how do we know? the total mechanical energy is conserved. At x = 1 m the kinetic energy is K 1 = 1 2 mv 1 2 = 1 2 (2.5 kg)(2.0 m/s)2 = 5.0 J and the potential energy is U 1 = 3( 1) 2 2( 1) 3 J = 5.0 J. So the total energy at x = 1 m is E mech = 10.0 J. The kinetic energy at x = 1 m is K 2 = E mech U 2 = 10.0 J 3(1) 2 2(1) 3 J = 9.0 J. v 2 = 2K 2 m = 2.68 m/s. = 2(9.0 J) 2.5 kg

8 Question 7.5: A pendulum consists of a 2.0 kg mass DISCUSSION PROBLEM Two identical balls roll along similar tracks the only difference is that track B has a small depression in it but the overall distances the balls travel from start to finish are the same. Since the balls fall the same overall heights (h) conservation of mechanical energy tells us that ignoring friction their speeds are the same at the end points. But if it s a race which one (if either) reaches the finish line first? hanging vertically from a string of negligible mass with a length of 3.0 m. The mass is struck horizontally so it has an initial horizontal velocity of 4.5 m/s. At the point where the string makes an angle of 30! with the vertical what is (a) the gravitational potential energy of the mass (b) the speed of the mass and (c) the tension in the string? (d) What is the angle of the string with the vertical when the mass is at its greatest height?

9 (b) Since energy is conserved K 2 + U 2 = K 1 + U 1 1 i.e. 2 mv2 2 + mgh = 1 2 mv2 1. Re-arranging we get v 2 2 = v1 2 2gh. v 2 2 = (4.5 m/s) 2 2(9.81 m/s 2 )(3.0 m)(1 cos30! ) The mechanical energy at all points of the swing is E = K + U. Take the zero of potential energy at i.e. put U 1 = 0. Then at K 1 = 1 2 mv2 1 : U 1 = 0. (c) = (m/s) 2 i.e. v 2 = 3.52 m/s. at K 2 = 1 2 mv2 2 : U 2 = mgh. (a) From the figure we deduce that h = l lcos30 ". U 2 = mg(l lcos30 " ) = mgl(1 cos30 " ) = (2.0 kg)(9.81 m/s 2 )(3.0 m)(1 cos30 " ) = 7.89 J. Identify all the forces acting on the mass at : T mgcos30! = mv 2 2 l

10 T = m v 2 2 l + gcos30" (3.52 m/s) 2 = (2.0 kg) + (9.81 m/s 2 )cos30 " 3.0m = 25.3 N. (d) At maximum height: K = 0 so U = K 1 i.e. mgh max = mgl(1 cosθ max ) = 1 2 mv 1 2. cosθ max = 1 v 1 2 2gl = 1 (4.5 m/s) 2 2(9.81 m/s 2 )(3.0 m) i.e. θ max = cos 1 (0.656) = 49.0 ". = Question 7.6: A physics student with mass 80 kg does a bungee jump from a platform 100 m above the ground. If the rubber rope has an un-stretched length of 50 m and spring constant k = 200 N/m (a) how far above the ground was the student at his lowest point? (b) How far above the ground did he achieve his greatest speed? (c) What was his greatest speed?

11 The solutions are y! = m and y! = m. appropriate solution is clearly y! = m. The Set the zero of gravitational potential energy at the ground. The total mechanical energy is therefore J which is conserved throughout the jump. At his lowest point (b) To find the position where the jumper s speed is greatest we need an expression for the kinetic energy (K) in dk terms of the vertical position (y) and set to find dy = 0 where K is a maximum. K kl2 + mgy " = J but K = 0 and l = 50 y ". 100(50 y! ) (9.81)y! = i.e. 100y! 2 ( )y! + ( ) = 0. y! y! = 0 i.e. y! = ± (92.15)2 4(1715.2) 2. If y is the height above the ground where the speed is greatest and l is the amount the rope has stretched then since mechanical energy is conserved at the fastest point K k l 2 + mgy = J.

12 Substituting for l = 50 y Check this is a maximum: we get K +100(50 y) y = i.e. K = (50 - y) y = y 100y 2 so dk = y. dy dk The extremum occurs when dy = 0 i.e. at y = = m. 200 d 2 K = 200 i.e. <0. 2 dy (c) K max = (50 y) 2 when y = m. K max = J = 1 2 mv 2 max i.e. v max = 2(40780 J) 80 kg = 31.9 m/s. In part (a) we found two possible values for the lowest point y! = m and y! = m. We chose the latter but what does the former correspond to? Note that both solutions correspond to positions where the kinetic energy of the jumper is zero. Consider the rope as a vertical spring and the jumper is the mass at the end of the spring. When released the mass oscillates up and down. The solution y! = m is the lowest point of the oscillation and the solution y! = m is the highest point of the oscillation. The mid-point at y = m is where the kinetic energy of the mass is greatest.

13 When non-conservative forces are involved such as kinetic friction deformation chemical reactions etc. we need a more general statement of the conservation of energy which involves the total energy of a system. Take for example a bike rider. The total energy of the system includes the mechanical energy the chemical energy of the rider friction etc. so E system = i E i = E mech + E chem + Providing there is no input to the system from external (e.g. wind a push) then E system = constant i.e. ΔE system = 0. So an increase in one form of energy (e.g. mechanical) is compensated by a decrease in another form (e.g. energy stored by the rider). Question 7.7: A block of mass 6.0 kg rests on an inclined plane. The coefficient of static friction between the block and the surface of the plane is A gradually increasing force is pulling down on the spring which has a spring constant of k = 200 N/m.. If the angle of the incline is θ = 40! what is the elastic potential energy of the spring when the block just begins to move?

14 The potential energy stored in the spring when the block is about to move is U = 1 where x is the extension of 2 kx2 the spring. Just before the block moves y F N F x = F s f s mgsinθ = 0 and F y = F N mgcosθ = 0 f s mg θ F s x Substituting for F s and f s we get kx µ s mgcosθ mgsinθ = 0 i.e. x = mg(sinθ + µ s cosθ). k Substituting the given values x = (6.0 kg)(9.81 m/s2 )(sin 40! cos40! ) 200 N/m = m. U = 1 2 kx2 = 1 2 (200 N/m)(0.268 m)2 = 7.18 J. where f s = µ s F N. But what is F s the force exerted by the spring? We know that U = 1 so that the magnitude of 2 kx2 the force associated with that potential function is F s = du dx = kx.

15 U G = 0 Question 7.8: A 2.0 kg block slides down a frictionless curved ramp starting from rest at a height of 3.0 m. The block them slides 9.0 m on a rough horizontal surface before coming to rest. (a) What is the speed of the block at the bottom of the ramp? (b) How much energy is dissipated by friction? (c) What is the coefficient of kinetic friction between the block and the horizontal surface? (a) From position to position ΔK = ΔU G i.e. (K 2 + U 2 ) = (K 1 + U 1 ). (K 2 K 1 ) = (U 1 U 2 ) i.e. K 2 = mgh = 1 2 mv 2 2. v 2 = 2gh = 2(9.81 m/s 2 )(3.0 m) = 7.67 m/s. (b) The energy dissipated by friction is responsible for changing the thermal energy of the system. The total energy is E sys = K + U G + U therm. With no external sources ΔE sys = 0. ΔK + ΔU G + ΔE therm = 0

16 But on the horizontal section i.e. ΔE therm = ΔK ΔU G. ΔU G = 0 and ΔK = K f K 2 = mv 2 2. ΔE therm = 1 2 mv 2 2 = 58.9 J. (c) The work done by friction (resulting in W f = f k Δx = µ k mgδx = ΔE therm ΔE therm ) is i.e. µ k = ΔE therm mgδx = 58.9 J (2.0 kg)(9.81 m/s 2 )(9.0 m) = Question 7.9 : A fireman slides down a pole at the fire station. If his speed of descent is constant which one of the following choices best explains the situation? A: U G K : E mech is not conserved but E sys is. B: U G E therm : both E mech and E sys are conserved. C: U G E therm : E mech is not conserved but E sys is. D: K E therm : E mech is not conserved but E sys is. E: K E therm : neither E mech nor E sys is conserved. U G is gravitational potential energy E therm is thermal energy produced by friction K is kinetic energy Emech is mechanical energy and system. E sys is the total energy of the

17 There are no external influences so E sys = K + U G + U therm = constant. ΔE sys = ΔK + ΔU G + ΔU therm = 0. But ΔK = 0. ΔU G = ΔU therm i.e. the loss in gravitational potential energy is converted to thermal energy through friction. Also since K is constant mechanical energy is not conserved. Therefore U G E therm : E mech is not conserved but E sys is. So the answer is C. Question 7.10: A block of mass 2.4 kg is dropped onto a spring with a spring constant k = 400 N/m from a height of 5.0 m above the top of the spring. (a) What is the maximum kinetic energy of the block. (b) What is the maximum compression of the spring? This is an interesting problem as I will use different positions for the zero of gravitational potential energy in parts (a) and (b).

18 But K 2 = mg ((5.0 m) + Δy) 1 2 k(δy)2 = (2.4 kg)(9.81 m/s 2 )(5.059 m) 1 (400 N/m)(0.059 m)2 2 = J. (b) Now we put U G = 0 at position. (a) Put U G = 0 at position where the kinetic energy is a maximum. Then conservation of energy gives K 1 + U G1 + U E1 = K 2 + U G2 + U E2 i.e. mg ((5.0 m) + Δy) = K k(δy)2. K 2 = mg ((5.0 m) + Δy) 1 2 k(δy)2. For maximum kinetic energy dk 2 = 0 = mg kδy d(δy) i.e. Δy = mg k = (2.4 kg)(9.81 m/s2 ) = m. 400 N/m Conservation of energy gives K 1 + U G1 + U E1 = K 3 + U G3 + U E3 i.e. mg ((5.0 m) + y! ) = 1 2 ky! y! y! = 0

19 i.e. y! = ± (23.45)2 + 4(200)(117.7). 2(200) y! = m or y! = m. Clearly the former is the physical meaningful solution and represents the maximum compression of the spring.