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Lesson 35 – Hooke’s Law & Elastic Potential Energy

Lesson 35 – Hooke’s Law & Elastic Potential Energy. By: Fernando Morales September 17, 2013. Learning Goals. Derive the work kinetic energy theorem Solve problems using the work kinetic energy theorem Derive the work-gravitational potential energy theorem

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Lesson 35 – Hooke’s Law & Elastic Potential Energy

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  1. Lesson 35 – Hooke’s Law & Elastic Potential Energy By: Fernando Morales September 17, 2013

  2. Learning Goals • Derive the work kinetic energy theorem • Solve problems using the work kinetic energy theorem • Derive the work-gravitational potential energy theorem • Investigate how the change of reference points affects the calculations of potential energy

  3. Work done by a force Work done is the area under a F|| vs d curve For those with calculus, this is W = F·dx For those without calculus, we can approximate the area for a changing force with rectangles

  4. 3 2 1 0 -1 d (m) 1 2 3 F (N) X6 - 3 find the work done EXAMPLE +3 J +2 J -1 J Solution The area under the curve is 4 Nm = 4 J; areas above the axis are +ve and areas below the axis are negative

  5. Work done by spring, Potential Energy A spring (a) stores energy when compressed (b) does work when released (c)

  6. Work done by a spring (a)Relaxed spring (b) Spring stretched by force FP (+ve) the spring pulls back with force FS = -kx, -ve since x is +ve (c) Spring compressed by FP (-ve now) the spring pushes back with FS = -kx, now FS is +ve since x is –ve

  7. Spring Force and Work Done Force increases linearly as a spring is compressed Area under curve = work done Wext = kx² = EPE (EPE is Elastic Potential Energy)

  8. X6 – Spring Work done by a spring A spring with constant 1000 N/m is compressed 14 cm. Find the force applied and the work done EXAMPLE Solution Since Fa = kx, the applied force is (1000 N/m)(0.14 m) = 140 N The work done is kx² = (1000 N/m)(0.14 m)² = 9.8 J

  9. m s kx1² m Dart gun A dart with mass 0.100 kg is loaded; the spring constant is 250 N/m. The spring is compressed 6.0 cm; if the dart separates from the spring when the spring is at its normal length (x = 0) what will the dart speed be? EXAMPLE X6 - 12 Solution The only force horizontally is the spring force; vertically gravity is balanced by the normal force exerted by the barrel (in flight the dart is a projectile) WNC = 0 = KE + PE (KE + PE)b= (KE + PE)a 0 + kx1² = mv2² + 0 v2² = v2 = 3.0

  10. X6 - 13 2 kinds of PE A mass m (2.60 kg) falls from rest 55.0 cm & compresses a spring Y = 15.0 cm. Find the spring constant. Assume the spring has negligible mass and take the 0 elevation at the top of the uncompressed spring. EXAMPLE Solution y +ve up I: ball drops 55.0 cm TME1 = TME2 mv1² + mgy1 = mv2² + mgy2 0+ mgh = mv2² + 0 v2 = √2gh v2 = 3.28m/s

  11. 2 kinds of PE A mass m (2.60 kg) falls from rest 55.0 cm & compresses a spring Y = 15.0 cm. Find the spring constant. Assume the spring has negligible mass and take the 0 elevation at the top of the uncompressed spring. EXAMPLE X6 - 13 Solution y +ve up I: ball drops 55.0 cm v2 = 3.28m/s II: spring compresses 15.0 cm k = 2(mv2² + mgY) TME2 = TME3 Y² mv2² + mgy2 + kx2² = mv3² + mgy3 + ky3² k = 1580 N/m mv2² + 0 + 0= 0 - mgY+ kY²

  12. 2 kinds of PE alternate method EXAMPLE X6 - 13 Solution y +ve up spring compresses 15.0 cm after ball drops 55.0 cm TME1= TME3 mv1² + mgy1 + ky2² = mv3² + mgy3 + ky3² 0+ mgh + 0= 0 - mgY+ kY² k = 2mg(h+ Y) Y² k = 1580 N/m

  13. Bungee Jump Dave falls 15.0 m before the bungee cord begins to stretch. His mass is 75 kg and the cord obeys Hooke’s Law (F = -kx); k = 50 N/m. How far below the bridge will the jumper fall? (Ignore mass of cord and air resistance) EXAMPLE 2mg 2mg k k X6 - 13 Solution Cannot be solved with Kinematic Equations as the acceleration changes – must use Energy Conservation Dave starts with 0 KE, max GPE, 0 EPE (a); at (b) he has some KE, some GPE and 0 EPE; at (c) all energy is EPE Set low point as y = 0  h = 15 m + y Energy is conserved (no air friction) (y)² – y – (15) = 0 KEb + PEb = KEa + PEa y = 40 m or -11 m 0+ mg(15 + y) = 0 + k(y)² Dave falls 55 m total This is a quadratic!

  14. X6 - 13 m = 25 g r = 20 cm h = ? the Loop I want to release a ball and have it go around the loop shown and come out again. What is the minimum height from which I can release the ball? EXAMPLE Solution need minimum speed at top of circle  no normal force between loop and ball  only mg acting as Fc At top, mg acts down as the centripetal force mv²/r = mg  v² = rg & TME = KE + PE = m(rg) + mg(2r) = 2.5mrg So mgh = 2.5mrg  h = 2.5 r as a minimum, or 50 cm here

  15. X2 - 3 k2 k1 > k2 k1 > k2 k2 EXAMPLE Two springs have constants k1 > k2, so spring 1 is stiffer than spring 2. On which spring is more work done if (a) the force applied is the same (b) the distance stretched is the same • Consider graphs of each case. • Same force  greater stretch for weaker spring and work done (area beneath curve) is greater • Same distance  less force for weaker spring and work done (area) is less for weaker spring

  16. Required Before Next Class • Read Section 5.1from Nelson Textbook • Finish Previous Homework

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