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