Work Done = Force x Distance 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 But a = F m v2 = u2 + 2as s = v2 2 F m W = F x mv2 2F v2 = 2as W = 1 2 mv2 s = v2 2a s = mv2 2F All Work Done = Ek = 1 2 mv2

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 W = mgh mg

i) Work against gravity: W = F x d W = mgsinΘ x d W = 1 x g x 7 25 x 5 Example 1 What work is done moving a mass of 1kg steadily up slope of length 5m, given that the sin Θ = 7 25 and μ = 5 12 (i) against gravity (ii) against friction i) Work against gravity: W = F x d Θ 7 25 W = mgsinΘ x d 1kg 5m Θ W = 1 x g x 7 25 x 5 W = g x 7 5 252 −72 =24 P W = 13.7J Forces are in equilibrium FR mgcosΘ ii) Work against friction: mg P = FR + mgsinΘ W = μmgcosΘ x d mgsinΘ P = μR + mgsinΘ W = 5 12 x 1 x g x 24 25 x 5 P = μmgcosΘ + mgsinΘ W = 2g = 19.6J friction gravity

Example 2 A 2kg mass falls vertically past point A at 1ms-1 and point B at 4ms-1 in a viscous liquid, with a constant resistance force of 9.6N. Find the distance AB F = mg – 9.6 v2 = u2 + 2as A B 1ms-1 4ms-1 mg 9.6N F = 10N 42 = 12 + 2 x 5 x s a = F m 16 - 1 = 10s a = 10 2 = 5 ms-2 s = 1.5m

An alternative solution uses Energy At A: Ek = 1 2 mv2 EP = mgh A B 1ms-1 4ms-1 mg 9.6N Ek = 1 2 x 2 x 12 EP = 19.6h Ek = 1J Ek = 1 2 mv2 At B: EP = mgh EP = 0J as h=0 Ek = 1 2 x 2 x 42 Ek = 16J 1 + 19.6h = 16 + Work done against resistance 1 + 19.6h = 16 + 9.6h 10h = 15 h = 1.5m

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: Force OR F Θ Θ Direction of travel d The amount of the Force going in the direction of travel is FcosΘ Work Done = F . d Work Done = FcosΘ x d

( ) ) ( ) ( Work Done = F . d Work Done = . = Work Done = 4x8 + (-3)x2 Example 1) A force of (4i - 3j) 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? ( ) 9 - 1 4 - 2 Work Done = F . d AB = Work Done = . ) ( 8 2 4 -3 ) ( 8 2 = Work Done = 4x8 + (-3)x2 Work Done = 26J

a = F m = 6i + 2j 2 = 3i + j ms-2 v = u + at vB = 0 + (3i + j) x 3 Example 2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B. a) acceleration b) vB c) Kinetic Energy at B a = F m = 6i + 2j 2 = 3i + j ms-2 v = u + at vB = 0 + (3i + j) x 3 vB = 9i + 3j ms-1 Ek = 1 2 x2x( 90 )2 |vB| = 92+32 |vB| = 90 Ek = 90J

Example 2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B. d) e) Position vector of B s = ut + 1 2 at2 AB 0 + 1 2 (3i + j) x 32 AB = 27 2 i + 9 2 j m AB = OA AB = OB - OB = AB + OA ) ( 27 2 9 2 ) ( 31 2 15 2 ) ( 2 3 + OB = = = 15.5i + 7.5j m

Example 2) At time t=0 a 2kg mass is at rest at A(2i + 3j). A constant force (6i + 2j)N causes an acceleration. 3 seconds later it passes B. f) Work done by the force Work Done = F . d Work Done = . ) 27 2 9 2 ( 6 2 Work Done = 6x 27 2 + 2 x 9 2 Work Done = 90J

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Homework: Sheet HW 6.1

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F = P v 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: F = P v

The maximum velocity occurs when the acceleration is zero. Example 1) The engine of a car is working at a constant rate of 8kW. The car has a mass of 1500kg is being driven along a level straight road against a constant resistive force of 425N. Find the acceleration of the car when its speed is 10ms-1 and it’s maximum speed. Fc = P v v = 10ms-1 425N The maximum velocity occurs when the acceleration is zero. 1500kg Fc = P v a = F m Therefore Fc must equal 425N Fc = 8000 10 a = 375 1500 vmax = P Fc Fc = 800N a = 0.25ms-2 vmax = 8000 425 Resultant Force = 800 - 425 vmax = 18.8ms-1 F = 375N

vmax occurs when the acceleration is zero. vmax = 21000 1060 Example 2) A car travelling along a level road against a resistive force of 700N, has a mass of 1800kg and a maximum speed of 30ms-1. If the car works at the same rate and the resistive force is unchanged find the maximum speed when ascending a slope of sin-1 1 49 . Fc = P v v = 30ms-1 Fc = P v R 700N 1800kg 1800kg 700N sin-1 1 49 P = Fvmax P = 700x30 1800g 1800gsinΘ P = 21 000W vmax = P Fc Fc = 700 + 1800gsinΘ vmax occurs when the acceleration is zero. vmax = 21000 1060 Fc = 1060N vmax = 19.8ms-1

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p = m x v j = Δp j = pF -pI also j = F x t 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 = pF -pI also j = F x t Momentum and Impulse can be scalar or vector quantities.

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

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

va = i - j ms-1 cosΘ = va.vb |va||vb| vb = 3i+2j ms-1 cosΘ = 1x3 +( −1 x2) ( 12+ −1 2 )( 32+22 ) cosΘ = 1 2 x 13 Θ = 78.690… Θ = 79o

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 8cm2 and the water comes out at 6ms-1. Find the magnitude of the force. Find volume of water Volume in 1 sec = CSA x v 8cm2 Volume in 1 sec = 8 100 2 x 6 v = 6ms-1 Volume in 1 sec = 3 625 m3

Find mass of water ρ of water is 1000kgm-3 m = V x ρ m = 3 625 x 1000 m = 4.8kg Find momentum p = m x v j = F x t p = 4.8 x 6 As p = j and t = 1sec p = 28.8 Ns 28.8 = F x 1 F = 28.8N