Work and Energy WorkEnergy Equivalence means work done

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Work and Energy Work-Energy Equivalence means. . . work done on a body is

Work and Energy Work-Energy Equivalence means. . . work done on a body is equal to energy gained and work done by a body equals energy lost. Work Done = Force x Distance We can use this to derive the formulas for Kinetic and Potential Energies. Consider the work done on a body at rest, with mass m, to accelerate it over a smooth surface for a distance d. W = F x d 2 2 v = u + 2 as v 2 = 2 as

Consider the work done on a body at rest, with mass m, to lift

Consider the work done on a body at rest, with mass m, to lift it through height h, without any acceleration or friction. W = F x d F = mg and d = h h mg W = mgh

 1 kg 5 m 7 Θ 1 kg P 25 Θ m n

1 kg 5 m 7 Θ 1 kg P 25 Θ m n gsi sΘ mg W = 13. 7 J co mg FR i) Work against gravity: W = F x d W = mgsinΘ x d Θ Forces are in equilibrium ii) Work against friction: P = FR + mgsinΘ W = μmgcosΘ x d P = μR + mgsinΘ P = μmgcosΘ + mgsinΘ friction gravity = 19. 6 J

Example 2 A 2 kg mass falls vertically past point A at 1 ms-1

Example 2 A 2 kg mass falls vertically past point A at 1 ms-1 and point B at 4 ms-1 in a viscous liquid, with a constant resistance force of 9. 6 N. Find the distance AB A 1 ms-1 F = mg – 9. 6 F = 10 N 9. 6 N B 4 ms-1 mg v 2 = u 2 + 2 as 42 = 12 + 2 x 5 x s 16 - 1 = 10 s = 5 ms-2 s = 1. 5 m

An alternative solution uses Energy At A: A 1 ms-1 Ek = 1 J

An alternative solution uses Energy At A: A 1 ms-1 Ek = 1 J At B: 9. 6 N B 4 ms-1 mg EP = 0 J as h=0 Ek = 16 J 1 + 19. 6 h = 16 + Work done against resistance 1 + 19. 6 h = 16 + 9. 6 h 10 h = 15 h = 1. 5 m

Work Done when Force Applied in different direction to Travel. When Force and direction

Work Done when Force Applied in different direction to Travel. When Force and direction are in the same direction we find the Work Done by multiplying them. If they are in different direction however: ce r Fo Θ Direction of travel The amount of the Force going in the direction of travel is FcosΘ Work Done = FcosΘ x d OR F Θ d Work Done = F. d

Example 1) A force of (4 i - 3 j) N moves an object

Example 1) A force of (4 i - 3 j) N moves an object from A (1, 2) to B(9, 4) along a smooth surface. Distances in metres. What is the work done by this force? ( ) =( 8 ) 2 AB = 9 - 1 4 - 2 Work Done = F. d ( )( ) 4 8 Work Done = . -3 2 Work Done = 4 x 8 + (-3)x 2 Work Done = 26 J

Example 2) At time t=0 a 2 kg mass is at rest at A(2

Example 2) At time t=0 a 2 kg mass is at rest at A(2 i + 3 j). A constant force (6 i + 2 j)N causes an acceleration. 3 seconds later it passes B. a) acceleration b) v. B v = u + at c) Kinetic Energy at B

Example 2) At time t=0 a 2 kg mass is at rest at A(2

Example 2) At time t=0 a 2 kg mass is at rest at A(2 i + 3 j). A constant force (6 i + 2 j)N causes an acceleration. 3 seconds later it passes B. d) AB AB = e) Position vector of B AB = OB - OA OB = AB + OA OB = ( ) + ( ) = 15. 5 i + 7. 5 j m 2 3

Example 2) At time t=0 a 2 kg mass is at rest at A(2

Example 2) At time t=0 a 2 kg mass is at rest at A(2 i + 3 j). A constant force (6 i + 2 j)N causes an acceleration. 3 seconds later it passes B. f) Work done by the force Work Done = F. d ( )( ) 6 Work Done = . 2 Work Done = 90 J

Page 265 Ex 11 E Q 1 -4 Q 8 -24 (even only)

Page 265 Ex 11 E Q 1 -4 Q 8 -24 (even only)

Homework: Sheet HW 6. 1

Homework: Sheet HW 6. 1

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Vehicles in Motion A vehicle travelling along a road has a forward force F

Vehicles in Motion A vehicle travelling along a road has a forward force F acting on it which is produced by an engine working at a constant rate of P watts. Power is the measure of the rate at which work is being done. Power = Work Done per second P = Force x distance moved per second P = Force x speed P = Fv Therefore the Force produced by an engine can be given by:

Example 1) The engine of a car is working at a constant rate of

Example 1) The engine of a car is working at a constant rate of 8 k. W. The car has a mass of 1500 kg is being driven along a level straight road against a constant resistive force of 425 N. Find the acceleration of the car when its speed is 10 ms-1 and it’s maximum speed. 425 N v = 10 ms-1 The maximum velocity occurs when the acceleration is zero. 1500 kg Fc = 800 N a = 0. 25 ms-2 Resultant Force = 800 - 425 F = 375 N Therefore Fc must equal 425 N vmax = 18. 8 ms-1

 v = 30 ms-1 R 700 N 1800 kg P = 700 x

v = 30 ms-1 R 700 N 1800 kg P = 700 x 30 P = 21 000 W vmax occurs when the acceleration is zero. 1800 g P = Fvmax g 00 k N 8 0 1 70 sin g 0 0 18 Fc = 700 + 1800 gsinΘ Fc = 1060 N Θ vmax = 19. 8 ms-1

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Page 271 Ex 11 G

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Momentum and Impulse Momentum, often given the symbol p, is the product of the

Momentum and Impulse Momentum, often given the symbol p, is the product of the mass and velocity of an object. It has the units Newton-seconds (Ns). In a closed system the total momentum always remains constant. Impulse is the difference in momentum caused by a Force acting on the object(s) over a time t. This is often given the symbol j. p = m x v j = Δp j = p. F -p. I also j = F x t Momentum and Impulse can be scalar or vector quantities.

Example 1) A 1 kg mass with velocity (2 i+3 j) ms-1 meets a

Example 1) A 1 kg mass with velocity (2 i+3 j) ms-1 meets a 2 kg mass with v = (-2 i+5 j) ms-1 The 2 masses collide and coalesce (stick together) What is their final speed and direction? Before Collision After Collision mt = 3 kg vt = ? ms-1 ma = 1 kg mb = 2 kg va = 2 i+3 j ms-1 vb = -2 i+5 j ms-1 p. I = ma x va + mb x vb p. I = 1(2 i+3 j) + 2(-2 i+5 j) p. I = -2 i + 13 j Ns pf = mt x vt pf = 3 vt = -2 i + 13 j

Example 2) A blue ball with velocity (4 i+j) ms-1 hits a stationary white

Example 2) A blue ball with velocity (4 i+j) ms-1 hits a stationary white ball of the same mass which then moves off with velocity v = (3 i+2 j) ms-1. a) What is the final velocity of the blue ball? b) What is the angle between the two final velocities? Before Collision ma = mkg mb = mkg ua = 4 i+j ms-1 ub = 0 ms-1 p. I = ma x ua + mb x ub p. I = m(4 i+j) + 0 p. I = 4 mi + mj Ns After Collision ma = mkg mb = mkg va = ? ms-1 vb = 3 i+2 j ms-1 pf = ma x va + mb x vb pf = mva + m(3 i+2 j) 4 mi + mj = mva + m(3 i+2 j) 4 i + j = va + 3 i+2 j va = i - j ms-1

va = i - j ms-1 vb = 3 i+2 j ms-1 Θ =

va = i - j ms-1 vb = 3 i+2 j ms-1 Θ = 78. 690… Θ = 79 o

Force on a Surface When a jet of water hits a wall and does

Force on a Surface When a jet of water hits a wall and does not rebound the loss of momentum is equal to the impulse from the wall. If we know how much water is hitting the wall we can find the momentum destroyed. Example: Water comes out of a horizontal pipe and hits a wall without rebounding. The CSA of the pipe is 8 cm 2 and the water comes out at 6 ms-1. Find the magnitude of the force. 1) Find volume of water Volume in 1 sec = CSA x v 8 cm 2 v = 6 ms-1

2) Find mass of water ρ of water is 1000 kgm-3 m = V

2) Find mass of water ρ of water is 1000 kgm-3 m = V x ρ 3) Find momentum p = m x v p = 4. 8 x 6 p = 28. 8 Ns j = F x t As p = j and t = 1 sec 28. 8 = F x 1 F = 28. 8 N