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![Solution
Work Done by spring
Work Done by spring
= F(x).dx
= -kx.dx
= -1
/2
kx2
= -2.5 [0.22
– 0.12
]
= -2.5 [0.04 – 0.01]
= -0.075 J](https://image.slidesharecdn.com/aworkenergyandpower-170330045200/75/A-work-energy-and-power-23-2048.jpg)
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A work, energy and power
- 1.
- 2. In life, thereare many forms of work
- 3. Work in Physics Inphysics, ‘work’ has a very specific meaning. Only one of the previous pictures shows ‘work’ in a physics sense Can you guess which one it is?
- 4. Definition of work Workis defined as the a FORCE which is applied over a DISTANCE. W = Fx Note that more work is done when … … a greater force is used … the force acts over a greater distance
- 5. What is NOTwork? Note that work is not done … … by holding an object in the same position … when a ball is kicked, (since the force is instantaneous)
- 6. Work can onlybe done while the force continues to act on the object over this distance.
- 7. The Units ofWork Work is a SCALAR unit The unit of work is JOULES (note this is the unit of energy) The fundamental units of work are … Nm = kg.m^2 s^2
- 8. Problem • What isthe work done by the forklift truck earlier?
- 9. When the forceis not parallel to the direction of motion. T Direction
- 10. When the forceis not parallel to the direction of motion. T T.cos(q) q
- 11. When the forceis not parallel to the direction of motion Now, Work Done … W = (F.cosq)x = Fx.cos(q) Note that any forces perpendicular to the direction of motions will not do any work on the object.
- 12. Negative Work Since: W= Fx.cos(q) And we know that cos(q) is .. positive from 0 to 90 deg. negative from 90 to 180 deg. Then when a force acts in the OPPOSITE direction of it’s motion, the work done is NEGATIVE
- 13. Example When the weightis RAISED… The man’s UPWARD force .. does POSITIVE work, Gravity’s DOWNWARD force .. does NEGATIVE work When the weight is LOWERED … The man’s UPWARD force.. does NEGATIVE work Gravity’s DOWNWARD force.. does POSITIVE work
- 14. Problem 1. A leafof mass ‘m’ falls from the top of a tree of height ‘h’. What is the work done by gravity on the leaf? 2. A second leaf of the same mass fall from the same height, but a breeze blows it so that it falls with an angle of ‘q’ degrees to the vertical. What is the work done by gravity on this leaf?
- 15. Solutions 1. W =Fx =mgh 1. W = (F.cosq)x = mg(cosq) x h/(cosq) = mgh q h h(cosq)
- 16.
- 17. The Force –Energy Theorem
- 18. Variable Force Sometimes, theforce applied on a body is not constant, but dependent upon it’s position Consider a mass on a spring
- 19. A mass ona Spring
- 20. W F1.dx +F2.dx + … + Fn.dx As dx 0, dx.ΣF F(x).dx F(x) = F x F = -kx -kx dx
- 21. For a variableforce Work … W = F(x).dx Note that for a constant force: F W = F.dx = Fx + d (as seen earlier) F x F x W W
- 22. Problem Calculate the generalformula for finding the work done of a mass on a spring What is the work done on a mass that is moved by a spring with constant k=5 a displacement of 10 to 20cm past its equilibrium?
- 23. Solution Work Done byspring Work Done by spring = F(x).dx = -kx.dx = -1 /2 kx2 = -2.5 [0.22 – 0.12 ] = -2.5 [0.04 – 0.01] = -0.075 J
- 24. Work and Energy F= ma … = m . dv/dt = m . dv . dx dx dt = mv . dv/dx Newton’s 2nd Law Chain rule
- 25. If W =F.dx And F = mv . dv/dx Then W = mv(dv/dx).dx = mv.dv Integrating between vi and vf … W = mv.dv = 1 /2 m(vf )2 – 1 /2 m(vi )2 = Kf – Ki = δK (The change in kinetic energy) The work done on an object is transformed into the object’s kinetic energy.
- 26. Problem Sheet 2 TheWork – Energy Theorem
- 27.
- 28. Mechanical systems, (eg:engines) are not so much limited by the work they can do, but the RATE at which they can do this work. The rate at which work is done is called: POWER
- 29. Since power isthe RATE of work, Average power … P = W/t The unit of power is JOULE PER SECOND, or … WATT
- 30. Note that ... aCONSTANT FORCE produces a constant acceleration Thus, the body will move a greater distance (x) each second Thus the work done each second will increase Thus the power will increase over time Therefore, a constant force does NOT produce a constant power.
- 31. Problem 1. A forceof 2N is applied to a stationary body of 1kg. How far will this body travel during each second for the first 3 seconds? 2. Calculate the average power applied during each second for the first 3 seconds 3. Find P(t) (Power as a function of time)
- 32. From F=ma, aconstant force of 2N produces a constant acceleration of 2m/s^2, (velocity increases by 2 seconds every second) Solution d(t) = 1/2at^2 + v.t + d d(1) = ½(2)(1)+0 =1m d(2) = ½(2)(1)+2 =3m d(3) = ½(2)(1)+4 =5m W = Fx W(1) = (2)(1) = 2J W(2) = (2)(3) = 6J W(3) = (2)(5) = 10J Since P = W/t, and the work is calculated over one second, the power over the first 3 seconds is also: 2W, 6W and 10W
- 33. d(t) = 1/2at^2 W(t)= Fd = F(1/2at^2) = 1/2Fat^2 P(t) = W/t = W/1 = 1/2Fat^2 P(t) = 0.5Fat^2 6W 10W 2W
- 34. Instantaneous Power In factinstantaneous power is usually calculated using calculus: P = dW/dt = F.cos(θ)(dx) =F.cos(θ) . dx/dt dt P = Fv.cos(θ)
- 35. Problem Sheet 3:Power 1. An elevator must lift 1000 kg a distance of 100m at a velocity of 4 m/s. What is the average power the elevator exerts during this trip? 2. When a 10 kg object which reaches terminal velocity at 100m/s, how much power does the air resistance exert on the object? 3. Derive, using the equation P = dW/dt, an expression for the power exerted by gravity on an object that is in free fall.
- 36. Summary Work done bya … Constant parallel force Constant non parallel force Variable parallel force Work - energy theorem Average power Instantaneous power For constant parallel force Constant non parallel force = Force x Distance = Fx.cos(q) = F(x).dx = dK = W/t = dW/dt = Force x Velocity = Fv.cos(q)
- 37. Points to remember •Work is only done when the Force continues to be applied over the distance • A constant force does not produce a constant power, but an increasing power
- 38. W = F1.dx+ F2.dx + F3.dx + = dx (F1 + F2 + F3 + …)F xdx F1 F2 F3 F4 F5
- 39. F(x) = F x F =-kx -kx