LAST NAME FIRST NAME DATE. Rotational Kinetic Energy. K = ½ I ω 2

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1 LAST NAME FIRST NAME DATE Work, Energy and Power CJ - Assignment The Conservation of Mechanical Energy Problems 3, 34, 38, 40 page 190 Work Kinetic Energy Rotational Kinetic Energy W = F d cosθ KE = ½ mv RKE = ½ I ω E = W = F d = F dcosθ K = ½ mv Strategy for Working Energy Problems K = ½ I ω Gravitational Potential Energy Short Distances: GPE = mg h Larger Distances: GPE = Gm 1m Short Distances: U g = mg y Larger Distances: U G = Gm 1m Elastic Potential Energy EPE = ½ k x U s = ½ k x 1. Draw a diagram of the situation including the important states.. Select the system and its starting (E1) and ending State (E). Write the Law of Conservation of Energy (W + E1 = E) 3. Write the Equation for each State, include all types of Energy (use the initials for the forms of energy) W + E1 = E W+ { KE 1 + RKE 1 + GPE 1 + EPE 1 } = { KE + RKE 1 + GPE + EPE } 4. Cross out values that are equal to zero to simplify the equation. 5. Substitute in the equations for the remaining forms of energy. W + { 1 mv Iω 1 + mgh mv 1 } = { 1 mv + 1 Iω + mgh + 1 mv } 6. Solve the remaining equation for the desired variable. 7. Plug and Chug STRATEGIES Mechanics problems (everything but Electricity problems) can be broken down into 3 strategies: 1. Equations of Motion (also called Kinematics)/ Newton s Laws. Momentum/Impulse/Collisions 3. Energy & Law of Conservation of Energy Rarely can more than one strategy be used. Before you start determine the best method. If it doesn t work, try another method. Below is an example of one time when either Kinematics or Energy can be used to solve the problem. You choose which is easier for you.

2 An elevator has an unknown mass. The elevator is at the top floor of the Sears Tower that is 440 meters high. If the cable holding the elevator were to snap, what velocity would it have when it hits the ground floor? (We must assume no friction is present) Solution Using Law of Conservation of Energy Solution Using Kinematic Equations (Diagram Left out ) (Diagram Left out ) The mass is not given so it must cancel. h 1 = 440 meters acceleration due to gravity = 9.8 m/s 10 m/s h = 0 meters All the energy in State 1 is gravitational energy; so all others can be cancelled. No work is included. All the energy in State is in the form of kinetic energy so all others can be cancelled. W + E 1 = E V i = 0 V f =? (Find this) acceleration due to gravity = 9.8 m/s 10 m/s Δy = h = 440 m Assume down is positive v f = v i + gx The first term cancels because initial velocity = 0, this leaves: W + {KE 1 + GPE 1 + EPE 1 } = {KE + GPE + EPE } v f = gh mgh 1 = 1 mv solving this for velocity gives v = gh v = gh Silly that these come out the same? Nah, Physics works! Ending with Work Starting with Work KE RKE GPE (short) GPE (long) EPE KE RKE GPE (short) GPE (long) EPE

3 3. A gymnast is swinging on a high bar. The distance between his waist and the bar is 1.1 m, as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all his mass is located at his waist, find his speed at the bottom of the swing. REASONING The only two forces that act on the gymnast are his weight and the force exerted on his hands by the high bar. The latter is the (nonconservative) reaction force to the force exerted on the bar by the gymnast, as predicted by Newton's third law. This force, however, does no work because it points perpendicular to the circular path of motion. Thus, W nc 0 J, and we can apply the principle of conservation of mechanical energy. SOLUTION The conservation principle gives 1 mv f mgh f 1 mv 0 mgh 0 E f Since the gymnast's speed is momentarily zero at the top of the swing, v 0 0 m/s. If we take h f 0 m at the bottom of the swing, then h 0 r, where r is the radius of the circular path followed by the gymnast's waist. Making these substitutions in the above expression and solving for v f, we obtain E 0 v f gh 0 g(r) (9.80 m/s )( 1.1 m) 6.6 m/s

4 34. The skateboarder in the drawing starts down the left side of the ramp with an initial speed of 5.4 m/s. If non-conservative forces, such as kinetic friction and air resistance, are negligible, what would be the height h of the highest point reached by the skateboarder on the right side of the ramp? The distance h in the drawing in the text is the difference between the skateboarder s final and initial heights (measured, for example, with respect to the ground), or h = h f h 0. The difference in the heights can be determined by using the conservation of mechanical energy. This conservation law is applicable because nonconservative forces are negligible, so the work done by them is zero (W nc = 0 J). Thus, the skateboarder s final total mechanical energy E f is equal to his initial total mechanical energy E 0 : 1 1 mvf mgh f mv 0 mgh0 E (6.9b) f E 0 Solving Equation 6.9b for h f h 0, we find that f v v hf h0 g h SOLUTION Using the fact that v 0 = 5.4 m/s and v f = 0 m/s (since the skateboarder comes to a momentary rest), the distance h is v 1 0 v f 5.4 m/s 0 m/s h 1.5 m g 9.80 m/s

5 38. The drawing shows a skateboarder moving at 5.4 m/s along a horizontal section of track that is slanted upward by 48 degrees above the horizontal at its end, which is 0.40 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track. REASONING To find the maximum height H above the end of the track we will analyze the projectile motion of the skateboarder after she leaves the track. For this analysis we will use the principle of conservation of mechanical energy, which applies because friction and air resistance are being ignored. In applying this principle to the projectile motion, however, we will need to know the speed of the skateboarder when she leaves the track. Therefore, we will begin by determining this speed, also using the conservation principle in the process. Our approach, then, uses the conservation principle twice. SOLUTION Applying the conservation of mechanical energy in the form of Equation 6.9b, we have mv mgh mv mgh 1 1 f f 0 0 Final mechanical energy Initial mechanical energy at end of track on flat part of track We designate the flat portion of the track as having a height h 0 = 0 m and note from the drawing that its end is at a height of h f = 0.40 m above the ground. Solving for the final speed at the end of the track gives v v g h h 5.4 m/s 9.80 m/s 0 m 0.40 m 4.6 m/s f 0 0 f This speed now becomes the initial speed v 0 = 4.6 m/s for the next application of the conservation principle. At the maximum height of her trajectory she is traveling horizontally with a speed v f that equals the horizontal component of her launch velocity. Thus, for the next application of the conservation principle v f = (4.6 m/s) cos 48º. Applying the conservation of mechanical energy again, we have mv mgh mv mgh 1 1 f f 0 0 Final mechanical energy Initial mechanical energy at maximum height of trajectory upon leaving the track Recognizing that h 0 = 0.40 m and h f = 0.40 m + H and solving for H give 0.40 m 0.40 m mv mg H mv mg 1 1 f 0 v v 4.6 m/s 4.6 m/s cos 48 0 f H 0.60 m g 9.80 m/s

6 40. A particle, starting from point A in the drawing, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 4.00 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A. REASONING AND SOLUTION The conservation of energy applied between point A and the top of the trajectory gives KE A + mgh A = mgh where h = 4.00 m. Rearranging, we find KE A = mg(h h A ) or A v g h h 9.80 m/s 4.00 m 3.00 m 4.43 m/s A