Slide 1 / 84 Slide 2 / 84 AP Physics C - Mechanics Energy Problem Solving Techniques 2015-12-03 www.njctl.org Slide 3 / 84 Slide 4 / 84 Table of Contents Click on the topic to go to that section Introduction · Gravitational Potential Energy Problem Solving · · GPE, KE and EPE Problem Solving Introduction · Conservation of Energy Problem Solving · The Spring and the Roller Coaster · Nonlinear Spring Potential Energy Graph Interpretation · Return to Table of Contents Slide 5 / 84 Slide 6 / 84 Introduction This is not a typical chapter presentation. It is a mix of step by step energy problem solutions, mixed in with free response and multiple choice formative assessment questions. Gravitational Potential The first four problems are non-calculus based. The remaining Energy Problem problems require calculus. Solving These can be done in class, led by the teacher, or they can be done by the students outside of class. Return to Table of Contents
Slide 7 / 84 Slide 8 / 84 GPE on an Incline GPE on an Incline Let's put together the concepts of The box starts with no velocity, two dimensional motion and and after it is pushed up a forces with GPE. displacement d, the block slides up, and it momentarily stops d d We'll use a box being pushed up before sliding back down. h h an incline. Does ΔGPE = mgd? # # It all depends on what we can measure. Assume it's easier to measure the displacement (d) the box travels. How do we find its change in GPE? Slide 9 / 84 Slide 10 / 84 GPE on an Incline GPE on an Incline No! The formula for ΔGPE was ΔGPE = mgh calculated from the work formula, and it assumed the gravitational When motion is along an incline, force (or the force that opposed it d d the change in height can be related to lift the object) was in the same h h to the distance traveled using direction of the object's motion. trigonometry. # # The gravitational force points sinθ = h/d down. Since work only includes the distance and force components h = dsin # that are in parallel, ΔGPE involves h and not d in the picture. How is h calculated from d? Slide 11 / 84 Slide 11 (Answer) / 84 1 A 5.0 kg block is at the top of a 6.0 m long frictionless 1 A 5.0 kg block is at the top of a 6.0 m long frictionless ramp, which is at an angle of 37 0 . What is the height of ramp, which is at an angle of 37 0 . What is the height of the ramp? the ramp? Answer 37 o 37 o [This object is a pull tab]
Slide 12 / 84 Slide 12 (Answer) / 84 2 The 5.0 kg block slides to the bottom of the 6.0 m long 2 The 5.0 kg block slides to the bottom of the 6.0 m long frictionless ramp, which is at an angle of 37 o . What is frictionless ramp, which is at an angle of 37 o . What is the change in its GPE? the change in its GPE? Answer 37 o 37 o [This object is a pull tab] Slide 13 / 84 Slide 14 / 84 GPE, KE and EPE Problem Solving Return to Table of Contents Slide 15 / 84 Slide 16 / 84 GPE, KE and EPE GPE, KE and EPE As with all energy, KE is a scalar. However, it relates No, it does not! directly to the velocity, and velocity is a vector. ΔGPE = mgh, where h is the vertical displacement (purely When we perform calculations of velocity from KE and along the y axis) that the object has moved. GPE, we need to be careful to relate the change in GPE only to the change in KE in the y direction - thus it only The incline displacement is not important - only the affects the velocity in the y direction. vertical displacement. Now, what about EPE? What about KE?
Slide 17 / 84 Slide 18 / 84 3 A projectile is fired at an angle of 45 0 . Which factor is GPE, KE and EPE required to calculate the maximum height the projectile reaches by using the Conservation of Total Mechanical In the case of EPE, the amount that the spring is Energy? compressed is the important variable - no trigonometry is required. A The total initial velocity of the projectile. Kinetic Energy will either use the vertical displacement an object covers (for its relationship to GPE) or the actual B The horizontal distance traveled by the projectile. displacement of the object from the spring's force (for its relationship to EPE). C The total distance traveled by the projectile. Let's work a couple of problems by using the Conservation D The x component of the velocity of the projectile. of TME to make this more clear. E The y component of the velocity of the projectile. Slide 18 (Answer) / 84 Slide 19 / 84 3 A projectile is fired at an angle of 45 0 . Which factor is 4 A spring launcher fires a marble at an angle of 52 to the required to calculate the maximum height the projectile horizontal. In calculating the energy available for reaches by using the Conservation of Total Mechanical transformation into GPE and KE, what value of x is used Students type their answers here Energy? in EPE = 1/2kx 2 ? A The total initial velocity of the projectile. B The horizontal distance traveled by the projectile. Answer E C The total distance traveled by the projectile. D The x component of the velocity of the projectile. E The y component of the velocity of the projectile. [This object is a pull tab] Slide 19 (Answer) / 84 Slide 20 / 84 4 A spring launcher fires a marble at an angle of 52 to the Energy Problem Solving horizontal. In calculating the energy available for transformation into GPE and KE, what value of x is used What is the final velocity of a box of mass 5.0 kg that slides 6.0 m Students type their answers here in EPE = 1/2kx 2 ? down a frictionless incline at an angle of 42 0 to the horizontal? v o = 0 The system will be the block. Answer The straight line displacement Since there is no friction, there of the spring, independent of are no external non conservative the x and y axes. forces and we can use the Conservation of Total Mechanical Energy. What types v = ? of energy are involved here? [This object is a pull tab]
Slide 21 / 84 Slide 22 / 84 Energy Problem Solving Only three types of energy have been discussed so far. And in this case, there is only GPE and KE: v o = 0 (KE + EPE +GPE) 0 = (KE + EPE + GPE) m = 5.0 kg becomes: d=6.0m (KE + GPE) 0 = (KE +GPE) v = ? θ=42 0 to streamline the notation, we'll assume that no subscript implies a final quantity (KE = KE f ) Slide 23 / 84 Slide 24 / 84 Energy Problem Solving Energy Problem Solving Consider the inclined plane problem that was just worked, h 0 =4.0m but add a spring at the bottom v o = 0 of the incline. m = 5.0 kg The spring will be compressed . a distance Δx and then released. Find the velocity of the box when d=6.0m it rises back to where it was first compressed - a height of Δh. v = ? What energies do we have to θ=42 0 consider? The velocity at the bottom of the incline is 8.9 m/s. Slide 25 / 84 Slide 26 / 84 Energy Problem Solving Energy Problem Solving Once compressed, the box has Before we proceed further with EPE, GPE and zero KE. the solution, think how hard this problem would be to solve When it loses touch with the without using Conservation of spring at Δh above its fully Energy. compressed point, it will have . . KE, GPE and zero EPE. Once the object is released and the spring starts moving away from its compressed state, the force is no longer constant - it (KE + EPE + GPE) 0 = (KPE + EPE + GPE) will require mathematical integration (calculus) to solve. becomes: Free body diagrams are not the best way to find the velocity of (EPE + GPE) 0 = (KE + GPE) the object.
Slide 27 / 84 Slide 28 / 84 Energy Problem Solving Energy Problem Solving Let's put in the equations and rearrange them to solve for v. We now have the equation for the velocity when the block rises a vertical displacement of Δh. . . But if we're only given Δx, how do we find Δh? Use trigonometry and recognize that Δh = Δxsinθ. Slide 29 / 84 Slide 29 (Answer) / 84 5 A box on an inclined plane is in contact with a spring. 5 A box on an inclined plane is in contact with a spring. The box is released, compressing the spring. For every The box is released, compressing the spring. For every increment Δx, the box moves down the incline, how increment Δx, the box moves down the incline, how much does its height, Δh, change? much does its height, Δh, change? A Δ x 2 A Δ x 2 B Δ xcos θ B Δ xcos θ Answer C C Δ xsin θ C Δ xsin θ D √Δ x D √Δ x E Δ xtan θ E Δ xtan θ [This object is a pull tab] Slide 30 / 84 Slide 30 (Answer) / 84 6 A box is held on top of a spring on an inclined plane of 6 A box is held on top of a spring on an inclined plane of angle θ = 31 0 . The box is released, compressing the angle θ = 31 0 . The box is released, compressing the spring. If the spring moves 7.0 m down the plane, how spring. If the spring moves 7.0 m down the plane, how much does its height, Δh, change? much does its height, Δh, change? A 7.0 m A 7.0 m B 6.5 m B 6.5 m Answer C 6.0 m C 6.0 m D D 3.6 m D 3.6 m E 3.0 m E 3.0 m [This object is a pull tab]
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