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Liceo Scientifico Isaac Newton Physics course Work Professor Serenella Iacino

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First of all in Physics work is always done by forces, in fact a force acting upon an object does work when the initial point of application of the force is displaced. Introduction

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Work F Fcos displacement s θ W = F cos θ s fig.1 θ F s = F s cos θ fig.2 F s θ s m joule = 1 Newton 1 meter = 1kg

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s fig.3 θ F work is positive s fig.4 F work is negative fig.5 work is zero θ < 90° θ s F θ θ > 90° θ = 90°

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When many forces act upon an object, we can obtain the total work by two equivalent methods: P F F N a θ fig.6 P N a a Fa θ F s W = P s cos 90° = 0 W = N s cos 90° = 0 W = F s cos 180° = -F s W = F s cos 1 method: st we can determine the work done by each force acting upon an object and then we add them to obtain the sum of work: W = -F s + F s cos sum a θ

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P F F a x y N θ fig.7 F x F y s a F = m a xx P = 0 x yy P =-P N = 0 N = N x y F =F cos F =F sen θ θ x F =-F ax F = 0 ay a a = a a = 0 y x o 2 method: nd we can determine the resultant of all forces acting upon an object and then we determine the work of the resultant: - - F + F cos sum F x F y 22 x 2 a θ F )(+() =F () == yy F = m a = 0 W = F s cos 0° = F s = -F s + F s cos sum a θ

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We can represent graphically on a Cartesian plane the work of a constant force F that is parallel to the displacement s and has the same direction, placing the displacement along the x axis and the force along the y axis. s F o F 1 1 s fig.8 W = F s = F s cos 0°= F s 11 11

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F s o fig.9 s 1 W = f(x) dx 1 s 0 f(x) Δs 1

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If an object, having mass m, falls from a height h, the angle between the gravity force and the displacement is zero, so the work done by this force is: fig.10 h Example 1 P W = P h = P h cos 0° = mgh

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During a football match, a ball of mass m is kicked up, reaching a height h and then falls to the ground. Calculate the work done by the gravity force P during the ascent and during the descent. ascent descent fig.11 h W = P s = P s cos 180° = -mgh W = P s = P s cos 0° = + mgh Example 2

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x y fig.12 F F F x y = F = 0 F el F The elastic force x F = Kx, where K is the spring constant F F F elx = -Kx = 0 el ely F elx = - KxHookes law is the relationship

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Compressed Spring F F F x y = -F = -Kx = 0 F F F elx ely = Kx = 0 el x y o fig.13 F F el

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F o y=-Kx x fig.14 Work Graphic representation of the elastic force x (-Kx) 2 2 W == 2 Kx s y= Kx x fig.15 F o s 2 W = x (+Kx) = Kx

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Kinetic Energy in Physics 2 from which we have then if i v = v + a t v = a t s = v t + at s = a t 2 1 W = F s = F s cos 0° = ma a t 1= m (a t) = m v Kinetic Energy ( K ) is energy of motion ( measured in joules)

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2 t = 2 i fi a vv fi a vv 2 fi a vv fi a vv - = 1 1 i v a fi a vv - fi 2 vv + = =s = = a v v t + i a t = fi a vv - fi 2 a vv - 22 W = F s = F s cos 0° = m a fi 2 a vv m v f 2 = 1 2 i 2 Variation of Kinetic Energy in Physics

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We observe that: 1.if the work is positive, then the kinetic energy increases when the object goes from the initial position to the final position; 2.if the work is negative, then the kinetic energy decreases; 3.if the work is zero, the kinetic energy doesnt change, so the velocity is constant. The work - Energy Theorem

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a man pushes a supermarket trolley of mass m = 10 kg, for two metres, applying a force F the magnitude of which is 100 N; the friction force between the floor and the trolley is 30 N. P F F N a fig.17 s W = W = 0 W = F s cos 180° = -F s = - 60,0 j W = F s cos 0° = 200 j F F P N a aa a Knowing that the force of gravity P and the normal force N also act upon the object, we draw the free-body diagram and we determine the work done by each force: Example: fig.16

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we have: sum m v f 2 = 1 2 i 2 W , = 10,0 ( v - 0 ) 2 f from which we have: v = f v = f = 5,29 m s

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THE END WORK

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