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Work

Learn about work in physics, how it's calculated, when it's zero, and its relation to energy through examples and explanations.

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Work

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  1. Work The work done on an object by a constant force is the product of the force that is parallel to the displacement (θ = 0°) times the displacement. Work is computed by W = F × Δx × cosθ where F is measured in N, Δx is the displacement measured in m and θ is the angle between the force and the displacement.

  2. The unit of work is 1N•1m = 1J where J is the work done measured in joules. Do not confuse a N•m which is also used to measure torque. Work is a scalar quantity meaning it has only direction.

  3. • When is there no work done in a “physics sense”? • When Δx = 0, W = 0 J in all cases. • No matter how long you hold a heavy package, if your displacement is zero, so is the work done. F You are applying a force but Δx = 0 which makes W = 0. Fw

  4. • F • What happens when you carry this heavy object a displacement of 2.0 m? You are exerting a force, you do have a displacement, but the angle between the applied force and the displacement is 90°, therefore the W = 0. Δx Fw

  5. What about an object undergoing uniform horizontal circular motion? You are exerting a force, you do have a displacement, but the angle is 90°, and the cos 90° = 0, therefore the W = 0. Fc Δx

  6. Work Problems How much work would you do if you pushed a 140 kg crate 12.7 m across a floor at a constant velocity if µ = 0.60? m = 140 kg Δx = 12.7 m µ = 0.60 FN Fy = 0 Fx = 0 F = Ff FN = Fw Ff F Fw

  7. µ = Ff/FN Ff = µFN = µFw = µmg Ff = 0.60 × 140 kg × 9.80 m/s2 = 820 N W = F × Δx × cosθ W = 820 N × 12.7 m × 1 = 1.0 x 104 J

  8. FN F FV θ Ff FH Fw A 60. kg crate is pulled 50. m along a horizontal surface by a constant force of 110. N which acts at an angle of 40° with the horizontal. The frictional force is 50. N. • What is the work done by each force acting on the crate. m = 60. kg F = 110. N Δx = 50. m θ = 40° Ff = 50. N

  9. W(Fw) = Fw × Δx cos 90° = 0 J W(Fv) = FV × Δx cos 90° = 0 J W(FN)= FN × Δx cos 90° = 0 J W(FH) = FH × Δx cos 40° W(FH) = 110. N × 50. m × cos 40° = 4.2 x 103 J

  10. W(Ff) = Ff × Δx cos 180° W(Ff) = 50. N × 50. m × cos 180° = -2.5 x 103 J (b) What is the net work done on the crate? WV = 0 J in a vertical direction. WT = FH + Ff + WV WT = 4.2 x 103 J – 2.5 x 103 J + 0 WT = 1.7 x 103 J

  11. Work-Energy Principle Energy is the ability to do work. Kinetic energy (translational energy) is the energy resulting from the motion of an object and is given by • KE = ½mv2 = 1 kg•1 m2/s2 = N•m = J • This is no coincidence that work and energy are measured in the same units.

  12. The work-energy principle states that the net work done on or by an object is equal to its change in kinetic energy. • W = ΔKE = ½mv2 • When ΔKE > 0, work is done on the object and its kinetic energy increases. • When ΔKE < 0, work is done by the object and its kinetic energy decreases.

  13. Gravitational potential energy is the stored energy an object has because of its position relative to the earth. The work-energy principle states that the net work done on or by an object is equal to its change in potential energy. • W = F × Δx × cosθ • W = ΔPEg = mgΔh = kg•m/s2•m = J

  14. Δh is the height measured from a convenient reference point. • The reference point can be anywhere because you are only interested in the change in energy, not the actual energy. • When ΔPEg > 0, work is done on the object and its potential energy increases. • When ΔPEg < 0, work is done by the object and its potential energy decreases.

  15. Work-Energy Principle Problem How much work is required to accelerate a 1250 kg car from 30. m/s to 40. m/s? m = 1250 kg vi = 30. m/s vf = 40. m/s W = ΔKE = ½mvf2 - ½mvi2 W = ½ × 1250 kg × (40. m/s)2 – ½ × 1250 kg × (30. m/s)2

  16. W = 4.4 x 105 J Because W > 0, work is done on the car increasing its kinetic energy.

  17. More Work-Energy Problems A 75 kg crate is pushed up a rough inclined plane by a force of 470 N which is parallel to the plane. The plane is 20. m long and makes an angle of 25° with the ground. • How much work is done by pushing the crate from the bottom to the top of the plane? m = 75 kg FT = 470 N l = 20. m θ = 25°

  18. FN FT • Fp θ Ff FN . θ Fw W = F × Δx × cosθ = 470 N × 20. m × 1 = 9.4 x 103 J

  19. (b) What is the change in potential energy of the crate? ΔPEg = mgh = 75 kg × 9.80 m/s2 × 20. m × sin 30° ΔPEg = 7.4 x 103 J

  20. (c) What is the work done against friction? Fp = Fwsinθ Fp = 75 kg × 9.80 m/s2 × .500 = 370 N FT = Fp + Ff Ff = 470 N – 370 N = 1.0 x 102 N W = Ff × Δx × cosθ

  21. W = 1.0 x 102 N × 20. m × -1 = -2.0 x 103 J It is important to realize that something is remaining constant in the problem. • That something is the sum of the ΔPEg (gravitational potential energy) and Wf (work done against friction) equals the work done in pushing the crate the entire length of the plane.

  22. A 70. g arrow is fired from a bow whose string exerts a force of 85 N on the arrow over a distance of 90. cm. What is the speed of the arrow as it leaves the bow? m = 70. g Δx = 90. cm F = 85 N W = ΔKE F × Δx × cosθ = ½mv2

  23. F × Δx × cosθ = ½mv2 v = (2 × F × Δx × cosθ/m)½ v = ((2 × 85 N × 90. cm × 1 m/100 cm × 1)/ 70. g × 1 kg/103 g))½ v = 47 m/s

  24. A 65 kg boy scout starts at an elevation of 1200. m and climbs to an elevation of 2700. m. • What is the boy scout’s change in potential energy? m = 65 kg g = 9.80 m/s2 hi = 1200. m hf = 2700. m ΔPEg = mgΔh = 65 kg × 9.80 m/s2 × (2700. m – 1200. m) = 9.6x 105J

  25. (b) What is the minimum work? W = ΔPEg = 9.6x 105J (c) Is it possible for the work required to be greater than the minimum amount of work? Yes, if friction was present, WT =Wf + ΔPEg

  26. Conservation of Energy Gravity is considered to be a conservative force. • When computing ΔPEg, the value doesnot depend on the path but only on the initial and final positions. • Sometimes gravity is referred to as a state function because it is independent of path.

  27. Friction is considered to be a non-conservative force or a dissipative force. • Because friction is dependent on path, it is sometimes referred to as a path function. • Whenever friction is present, there will always be a certain amount of mechanical energy that gets “wasted” or turned into heat.

  28. The amount of energy “wasted” due to friction does depend on the path taken. The conservation of mechanical energy holds true only if there are conservative forces acting on a system. • ΔE = 0 ΔKE + ΔPE = 0

  29. A Pendulum Problem A 0.57 kg pendulum bob at the end of a 0.90 m cord is displaced 0.10 m and then released. Using only energy considerations, determine the maximum speed of the bob. The center of gravity was raised 0.0050 m. m = 0.57 kg l = 0.90 m Δx = 0.10 m Δh = 0.0050 m g = 9.80 m/s2

  30. . Δh ΔE = 0 (KE + PEg)amp = (KE + PEg)eqpos 0 0

  31. (PEg)amp = (KE)eqpos mgΔh = ½mv2 . v = 2gΔh v = (2 × 9.80 m/s2 × 0.0050 m)½ v = 0.31 m/s

  32. W J P = = Δt s Power Power is the rate at which work is done or the rate at which energy is transformed into another form of mechanical energy. where P is the power measured in watts, W, W is the work in joules, J, and Δt is the time in seconds, s. W =

  33. Power is sometimes expressed in horsepower, 1 hp = 746 W. Power can also be expressed as W F × Δx × cosθ P = = F × v × cosθ = Δt Δt

  34. A Power Problem A 60. kg boy runs up a flight of stairs in 3.7 s. The vertical height of the stairs is 4.2 m. • What is the boy’s output power both in watts and horsepower? m = 60. kg Δt = 3.7 s Δh = 4.2 m ΔPEg mgΔh W P = = = Δt Δt Δt

  35. 60. kg × 9.80 m/s2 × 4.2 m P = 3.7 s 1 hp P = 670 W × = 0.90 hp 746 W E = P Δt . (b) How much energy is involved? E = 670 W × 3.7 s = 2.5 x 103 J

  36. Elastic Potential Energy To stretch or compress a spring an amount s from its equilibrium position requires a force F. Hooke’s Law is given by Fr = -ks where Fr is the restoring force in N, k is the spring constant in N/m, and s is the displacement of the spring from the equilibrium position.

  37. The negative sign indicates the restoring force of the spring. • The restoring force of the spring is always trying to restore the spring to its equilibrium position. • The greater the value of k, the stiffer the spring and the more difficult it is to stretch or compress it.

  38. Because Frα s, the more a spring is compressed or stretched, the greater the restoring force Fr. • W = ½•F•s•cosθ • The cos θ always equals 1 and Fr = ks which gives W = ΔPEe = ½ks2

  39. Applying the Conservation of Mechanical Energy to a spring gives • ΔE = 0 • (PEe + KE)amp = (PEe + KE)eqpos 0 0 PEe = KE ½ks2= ½mv2

  40. Spring Problem A spring stretches 0.150 m when a 0.300 kg mass is suspended from it. The spring is stretched an additional 0.100 m from this equilibrium position and released. 0.300 kg 0.150 m 0.300 kg 0.100 m

  41. mg Fw s s 0.300 kg × 9.80 m/s2 k = 19.6 N/m = 0.150 m • Determine the spring constant. m = 0.300 kg s = 0.150 m Δx = 0.100 m Fr = ks Fr = = k = s

  42. (b) What is the amplitude? The amplitude is 0.100 m which is the maximum displacement from the equilibrium position. (c) Determine the maximum velocity. (PEe + KE)eqpos = (PEe + KE)amp 0 0

  43. ½mv2 = ½kΔx2 v = (kΔx2/m)½ v = (19.6 N/m × (0.100 m)2/0.300 kg)½ v = 0.808 m/s in the direction of either amplitude

  44. (d) Determine the velocity when the mass is 0.050 m from the equilibrium position. (PEe + KE)amp = (PEe + KE)disp ½kΔx2 = ½ks2 + ½mv2 19.6 N/m × (0.100 m)2 = 19.6 N/m × (0.050 m)2 + 0.300 kg × v2 0

  45. v = 0.700 m/s in the direction of either amplitude (e) Fnet = ma a = Fnet/m = Fr/m = kΔx/m 19.6 N/m × 0.100 m a = = 6.53 m/s2 0.300 kg

  46. Wrap Up Questions A mass that is attached to the free end of a spring undergoes simple harmonic motion. Does the total energy change if the mass is doubled and the amplitude remains the same? The total energy is given by PEe = ½kxmax2, thereforechanging the mass has no effect on the total energy.

  47. Does the kinetic energy depend on mass? Yes, because kinetic energy, KE, is given by KE = ½mv2. If the net work done on an object is equal to zero, what is true about the object’s velocity? The velocity remains the same because W = ΔKE = 0 meaning that vf = vi.

  48. Why is the work done by a frictional force, Ff, is always negative? The frictional force always opposes the displacement of the object making the angle between the force and displacement 180°, and the cos 180° = -1.

  49. Can the average power ever equal the instantaneous power? Yes, if the power remains constant throughout a time interval.

  50. Using the work-energy principle, show why friction always reduces the kinetic object of an object. Because the angle between the displacement and the frictional force is always 180°, W = ΔKE = ½mvf2 - ½mvi2 < 0 because vf < vi.

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