Liceo Scientifico Isaac Newton Physics course Work Professor Serenella Iacino

Introduction 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.

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

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

When many forces act upon an object, we can obtain the total work by two equivalent methods: st 1 method: 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 = P ∙ s ∙ cos 90° = 0 W = N ∙ s ∙ cos 90° = 0 W = F ∙ s ∙ cos 180° = -F ∙ s W = F ∙ s ∙ cos → P → → N F → N → θ F a → F a a a → → θ P F → s W = -F s + F ∙ s ∙ cos θ fig.6 ∙ sum a

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

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. → → F → → F 1 W = F ∙ s = F ∙ s ∙ cos 0°= F ∙ s 1 1 1 1 o s s 1 fig.8

F f(x) s Δ s s o 1 1 fig.9 W = ∫ f(x) dx s 1

Example 1 → 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: P h fig.10 → → W = P ∙ h = P ∙ h ∙ cos 0° = mgh

Example 2 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. → W = P ∙ s = P ∙ s cos 180° = -mgh → → ascent h W = P ∙ s = P ∙ s cos 0° = + mgh → → descent fig.11

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

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

Work Graphic representation of the elastic force y= Kx x x o s s o y=-Kx fig.14 fig.15 x ∙ (-Kx) -1 Kx 2 x ∙ (+Kx) +1 = Kx 2 W = = W = 2 2 2 2

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

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

The work - Energy Theorem We observe that: if the work is positive, then the kinetic energy increases when the object goes from the initial position to the final position; if the work is negative, then the kinetic energy decreases; if the work is zero, the kinetic energy doesn’t change, so the velocity is constant.

Example: 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. → → 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: → → a → N W = W = 0 W = F ∙ s ∙ cos 180° = -F ∙ s = - 60,0 j W = F ∙ s ∙ cos 0° = 200 j → → F F a → → N P → → P a a → F a s fig.16 → fig.17 F

we have: 1 1 W m v - m v = 2 2 1 - 60,0 + 200 = ∙ 10,0 ∙ ( v - 0 ) 2 f i sum 2 2 1 - 60,0 + 200 = ∙ 10,0 ∙ ( v - 0 ) 2 2 f from which we have: 140 140 v = 2 v = = 5,29 m s f 5 f 5

WORK THE END