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Work and Energy

Work and Energy. Definition of Work (W) . Work is done when a force is exerted on an object AND the object moves in the direction of the force (or one of its components)!. Is Work done (to the object) when. when you push box across the floor?.

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Work and Energy

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  1. Work and Energy Definition of Work (W) Work is done when a force is exerted on an object AND the object moves in the direction of the force (or one of its components)! Is Work done (to the object) when... • when you push box across the floor? YES- applied force in direction of distance covered YES- F is up and so is ∆d • when you lift a briefcase? • by the normal force acting on a sliding block? NO- F is up, ∆d is across!

  2. The amount of work done depends directly upon the amount of force that is doing the work and how far the object is moved: units of work: W = F∆d or Fcosø(∆d) J = Nm A Joule (J) is the standard unit of work (as well as the standard unit of energy) and it equals Newtons times meters! • Work is the scalar product of two vectors! • Work has no direction, BUT work done opposite the motion is considered negative!

  3. When determining work, you need to consider which force: Fa does (+) work N Fa Gravity and friction do negative work: Fp Ff How much work is done by the normal force? ø Fn ø Fw 0 J N is 90˚ from ∆d

  4. The amount of work (energy expended) done in each case is the same– less force is used over a greater distance, but ultimately the same work is done!

  5. Calculate how much work is done by the applied force when a crate weighing 850.0 N is pushed up a 15.0˚ incline at constant speed for a distance of 10.0 m. The µ between the crate and the plane is .240. N Fa W = ? Fp Fw = 850.0 N Ff ø = 15.0˚ ø ∆d = 10.0 m Fn ø µ = .240 Fw

  6. Balanced Forces? Yes- constant speed! Fa = Ff + Fp N = Fn Fp = sinø(Fw) = sin(15.0˚)(850 N) W = Fa∆d 220 N Fn = cosø(Fw)= cos15.0˚(850 N) = 821 N Ff = µFn = .240(821 N) = 197 N Fa = Ff + Fp = 197 N + 220 N = 417 N = (417 N)(10.0 m) = 4170 J

  7. A girl pulls a wagon with constant velocity along a level path for a distance of 45.0 m. She pulls with 85.0 N of force along the handle which makes an angle of 20.0˚ with the horizontal. How much work does she do? N ∆d = 45.0 m Fa Fa = 85.0 N Ff Fv ø ø = 20.0˚ Fh W = ? Fw

  8. N + Fv = Fw Fh = Ff W = Fh∆d Fh = cosø(Fa) = cos(20.0˚)(85.0 N) = 79.9 N = (79.9 N)(45.0 m) = 3600 J

  9. 1) A 25.0 kg box slides at constant speed down a ramp of 15.0˚. How much work is done by gravity if the ramp is 8.00 m long? 507 J 2) A car traveling 20.0 m/s slams on the brakes and stops in a distance of 12.0 m. If the car weighs 14,100 N, how much work is done by friction in stopping the car? -288,000 J 3) A 245 N crate is pulled by a rope up a 20.0˚ incline at constant speed. How much work is done by the rope if the crate is pulled up 6.00 m and µ = .250. 848 J

  10. 4) A crate that weighs 250.0 N accelerates from rest down a 20.0˚ incline until it reaches 5.00 m/s at the bottom. How much work was done to the box by gravity if µ = .200? 708 J 5) A 200.0 N sled is accelerated from rest to 15.0 m/s in 20.0 s. If the rope used to pull the sled is tied horizontally, how much work is done by the rope to the sled during this 20.0 s if µ = .225 ? 9050 J 6) A man accelerates a 25.0 kg crate from rest up a 2.00 m ramp in 5.00 s. How much work does the man do if the ramp is at an incline of 15.0˚ with horizontal and µ = .250? 253 J

  11. Fat guy: 300 lbs Skinny guy: 150 lbs Who does more work going up these stairs if the fat guy takes 10 s and the skinny guy takes 5 s?

  12. Power (P) • does not refer to strength as is the common definition! • the timed rate at which work is done (or the rate at which energy is expended) • depends directly upon the work and inversely upon the time to do that work! W ∆t F∆d ∆t P = =

  13. units for power: J s one watt (w) equals one Joule (J) per second (s) = w Power can also be measured in Horsepower (Hp). 1 Hp = 746 w (25 w)(1 Hp/746 w) = .0335 Hp

  14. A 35.0 kg box falls off of a truck that is traveling at 25.0 m/s and slides to a stop over a distance of 140 m. How much power is generated by friction during this time? N m = 35.0 kg Ff Fw = 343 N v0 = 25.0 m/s v = 0 Fw ∆d = 140 m P = ? Ff = Fnet Fw = N

  15. P = Ff ∆d ∆t Ff = ma a = vf2 -vi2 2∆d = (35.0 kg)(-2.23 m/s2) = -78.1 N = 0 - (25.0 m/s)2 2(140 m) = (-78.1 N)(140 m) 11.2 s = -2.23 m/s2 = - 976 w ∆t = vf - vi a negative power means the work done is opposite of the motion! = 0 - 25.0 m/s -2.23 m/s2 = 11.2 s

  16. A 45.0 kg crate is accelerated from rest up a 20.0˚ incline at 2.00 m/s2. At what rate was the work done by the applied force if the ramp is 6.00 m long and it took 7.00 s to reach the top (µ = .230)? N Fa m = 45.0 kg vi = 0 Fp Ff ø = 20.0˚ a = 2.00 m/s2 P = ? ∆d = 6.00 m Fn ∆t = 7.00 s Fw µ = .230

  17. Fa > Fp + Ff N = Fn Fp = sinø(Fw) Fa = Fp + Ff + F = sin(20.0)(441 N) = 151 N P = Fa∆d ∆t N = Fn = cosø(Fw) = cos(20.0)(441) = 414 N Ff = µN = (.230)(414 N) = 95.3 N = 288 w Fnet = ma = 45.0 kg(2.00 m/s2) = 90.0 N Fa = 151 N + 95.3 N + 90.0 N = 336 N

  18. 1) A 4.00 m ramp is used to pull a 25.0 kg box up a 15.0˚ incline at constant speed. At what rate was the work done if it took 5.00 s to get the crate to the top and µ = .230? 2) A 44.0 kg crate is allowed to slide from rest down a 3.50 m ramp until it reaches the bottom of the 18.0˚ incline 2.64 s later. At what rate was the work done if µ = .200? 3) A 45.0 kg crate is pulled across a level floor (µ = .220) by a force of 155 N applied to a rope that is tied at a 25.0˚ to the horizontal. The crate is pulled from rest for 6.00 s. How much power is generated during this time?

  19. 4) A 452 N crate is pulled across a level floor at constant speed by a force of 135 N applied to a rope tied at a 30.0˚ angle to the horizontal. How much power is generated if the crate is pulled a distance of 12.0 m in 15.0 s? 5) How much work is done in accelerating a 35.0 kg crate at 2.00 m/s2 up a 16.0˚ incline if the incline is 5.00 m long and µ = .210? 6) A 34.5 kg crate falls off the back of a truck traveling 18.0 m/s and slides to a stop in 7.00 s. What is the average power generated by friction during this time?

  20. Energy general term that refers to the ability to do work! there are MANY ways to classify types of energy! the two most fundamental are: potential energy kinetic energy

  21. Potential Energy • stored up energy- the potential to do some kind of work • that potential can come from a variety of sources: gravitational PEpotential due to position and gravity elastic PEpotential due to a stretched or compressed elastic object chemical PEpotential due to chemical reaction nuclear PEpotential due to nuclear chain reaction

  22. Gravitational Potential Energy (Ep) • potential energy because gravity could move an object a certain distance • depends upon mass, gravity, and height • height is determined from a set reference point- and that point can change! units are Nm = J Ep = mg∆h = Fw∆h the standard unit of energy is the Joule (J)!

  23. How much potential energy would a 2.0 kg book have if it were held 1.0 m above a desktop that is 1.0 m above the floor that is 4.0 m above the ground? Ep = mg∆h relative to the desk: Ep = (2.0 kg)(9.80 m/s2)(1.0 m)= 20 J relative to the floor: Ep = (2.0 kg)(9.80 m/s2)(2.0 m) = 40 J relative to the ground: Ep = (2.0 kg)(9.80 m/s2)(6.0 m) = 120 J Unless specifically told otherwise, gravitational Ep is relative to the surface of the earth!

  24. Elastic Potential Energy (Ep) • potential energy due to a stretched or compressed elastic object - like a spring! • in order to stretch or compress a spring, the force applied must increase with the amount of stretch/compression of the spring. 2 N 1 cm 4 N 2 cm 3 cm 6 N 4.5 cm 9 N the relationship here is constant!

  25. The ratio of the amount of force needed to stretch (or compress) a spring a certain amount is called the force constant! (k) Hooke’s Law N m F ∆d k = units  If 50.0 N is used to stretch a spring a total distance of 10.0 cm: 50.0 N = 5.00 N = 25.0 N .100 m .0100 m .0500 m k = = 500 N/m

  26. If 40.0 N is used to stretch a spring 10.0 cm, what is the force constant? k = F ∆d = 40.0 N .100 m = 400 N/m How much work was done in stretching the spring? W = F∆d = (40.0 N)(.100 m) = 4.00 J WRONG!

  27. The 40.0 N of force was the final force reached, it increased steadily from 0 N. The average force applied then was : Fav = Fi + Ff 2 = 0 + 40.0 N 2 = 20.0 N With a spring, work is calculated using the average force (because the force constantly changes): W = Fav∆d = (20.0 N)(.100 m) = 2.00 J

  28. The work done in stretching the spring will equal the gain in elastic potential energy. Therefore the spring will gain 2 J of potential energy: So how do you find the PE of a spring? units  J Ep = .5k∆d2 • for the given spring: Ep= .5k∆d2 = .5(400 N/m)(.1 m)2 k = F ∆d = 40 N .1 m = 2 J = 400 N/m

  29. A weight of 5.5 N will compress a spring exactly 1.0 cm. How much potential energy will the spring have when it is compressed 4.2 cm? F = 5.5 N Ep = .5k∆x22 ∆x1 = .010 m .5(550 N/m)(.042 m)2 ∆x2 = .042 m .49 J Ep = ? = 5.5 N .010 m k = F ∆x1 = 550 N/m

  30. If an object is moving for any reason, then it is said to have kinetic energy. Kinetic energy (Ek) is directly related to the mass of the object as well as the square of the velocity! Ek = .5mv2 Law of Conservation of Energy In a closed system (no energy lost to forces such as friction), the total amount of energy (the KE and PE) must remain constant! Ep + Ek = Et ∆Ep + ∆Ek = 0 or

  31. Energy is never lost- only converted from one form to another! W = ∆K -> The Work-Energy Theorem

  32. A rock that weighs 1120 N is at the edge of a cliff that is 22.0 m high. How much Ep and Ek does the rock have at the top of the cliff? Fw = 1120 N Ep = mg∆h = (1120 N)(22.0 m) m = 114 kg = 24, 600 J ∆h = 22.0 m Ek = 0 J (it is not moving!) v = 0 For this system, the total energy (Et): Et = Ep + Ek = 24,600 J + 0 J = 24,600 J How much Ep and Ek does it have as it hits after falling to the ground?

  33. h = 0 Ep = mgh = 0J Ek = 24,600 J Et = Ep + Ek = 0 + 24,600 J = 24,600 J A check of this answer: vi = 0 Ek = .5mv2 = .5(114 kg)(20.8 m/s)2 ∆d = -22.0 m = 24,600 J a = -9.80 m/s2 Ek = ? vf = vi2 + 2a∆d = 0 + 2(-9.80 m/s2)(-22.0 m) = 20.8 m/s

  34. What speed will the object have when it has fallen half way? Solving using conservation of energy: Ek = .5 mv2 Et = 24,600 J Ek ∆h = 11.0 m v = .5m Ep = 12, 300 J Ek = 12, 300 J 12,300 J = v = ? .5(114 kg) = 14.7 m/s

  35. Checking as a freefall problem: vi = 0 vf = vi2+ 2a∆d ∆d = -11.0 m = 2 (-9.80 m/s2)(-11.0 m) a = -9.80 m/s2 vf = ? = 14.7 m/s What speed will the object have after it has fallen a distance of 6.7 m? ∆h = 22.0 - 6.7 m = 15.3 m Ep = mg∆h = (1120 N)(15.3 m) = 17,100 J

  36. Ek = Et - Ep = 24, 600 J - 17,100 J = 7500 J Ek 7500 J v = = 11.5 m/s = .5m .5(114 kg) A spring with a force constant of 1560 N/m is compressed 12.0 cm and used to launch a mass of 1.00 kg horizontally. With what speed will the mass leave the spring? k = 1560 N/m m = 1.00 kg v = ? ∆x = .120 m Fw= 9.80 N

  37. The Ep of the compressed spring will be turned into Ek of the launched object! Ep = .5k∆d2 Ek = .5mv2 = .5(1560 N/m)(.120 m)2 = 11.2 J Ek v= .5m 11.2 J .5(1.00 kg) = 4.73 m/s = How high above the vertically compressed spring will the object be launched? Ans: 1.14 m

  38. Solve using the conservation of energy: A) A spring is used to launch a 2.50 kg projectile with a horizontal speed of 15.0 m/s. If the spring is compressed 30.0 cm before the launch, what is the force constant of the spring? 6250 N/m B) A 125 kg rock is dropped from a cliff so that it hits the ground with a speed of 42.0 m/s. How high is the cliff from which it was dropped? 90.0 m

  39. 1) A bowstring of force constant 1230 N/m is stretched 30.0 cm. What is the weight of an arrow that is launched horizontally with a velocity of 25.0 m/s? (vertically?) 2) A boulder of mass 255 kg is sitting atop a hill that is 34.6 m high. The boulder rolls down that hill and over the next hill, which is only 18.9 m high. Assuming the boulder loses no energy to friction and such, what is the speed of the boulder as it passes over the next hill?

  40. 3) A spring of constant 1250 N/m is compressed 45.0 cm and used to launch a 5.00 kg box up in the air. If the coefficient of friction is .250, how far above the uncompressed spring will the box rise before stopping? 4) A 42.0 kg box slides from rest down a frictionless 25.0˚ incline. It accelerates for 3.48 s to reach the bottom of the incline where it encounters a spring bumper that is compressed 40.0 cm in stopping the box. What is the force constant of this spring?

  41. 5) A spring of force constant 655 N/m is used to launch a 5.25 kg mass from the edge of a frictionless table that is 1.20 m above the floor. How far horizontally away from the table will the mass land if the spring was compressed 42.0 cm before launching? 2.32 m

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