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Lavoro (Modulo 3)
L = F · S = FS cosθ
[L] = [F][S] = [N][m] = [m][J]
Ltot = l1 + l2 + l3 + ... lm = i=1∑n li
= F1S1 + F2S2 + F3S3 + ... + FmS = (F1 + F2 + [...] + Fm) S =
(i=1∑nF)
S = RS
L = FS cosθ ⇒ L = 0 ⇒se F = 0se cosθ = 0 ⇒ θ = π/2se S = 0
FS = Fx x + Fyŷ + Fzz
S = Sx + Syŷ + Szz
Calcolo del lavoro
dL = f(x) dx
L = ⌠x2x1 f(x) dx = ⌠x2x1 f(x) dx
A = L = ⌠x0 f(x) = ½ kx²
F = Fxŷ + Fyŷ + Fzz
dS = dx ŷ + dyŷ + dz z
L = ⌠x2x1 f(x) F fx dx + ⌠y2y1 fy dy + ⌠z2z1 fz dz
⌠x2x1 F f dx = ⌠t2t1 m dv dx = ⌠ux2ux1 m dx
½ m vx22 - vx12
dvx = dvx dt = dvx dt · dx dt = dx dt d(dx dt)
L = ½ m [(vx22 - vx12) + (vy22 - vy12) + (vz22 - vz12)] = ½ m [(vx22 + vy22 + vz22) - (vx12 +vy12 + vz12)]
L = ½ m (v22 - v12)
v2 = vx2 + vy2 + vz2
v22 = v12 + vxvx + vyvy + vzvz = vx2 + vy2 + vz2
L = ½ m v22 - ½ m v12
L = K2 - K1 = ΔK
Teorema dell’energia cinetica
K = ½ mv2
L = ΔK = ½ mvF2 - ½ mvi2
- se L > 0 → ΔK > 0 → KF > Ki → vF > vi
- se L = 0 → ΔK = 0 → K = Ki → vF = vi
- se L < 0 → ΔK < 0 → KF < Ki → vF < vi
Fd
N
mg
S
LT = μd · N · S
Fd
x
y
y1
y2
x
vg
LF = mgh cos(⊘) = mgh = ΔK = ½ mvF2 - ½ m vi2
Lp = mgh = ½ mvF2
vF2 = 2gh → vF = √2gh
- mgh = Lp = ½ mvF2- ½ mvi2
mgh = ½ mvi2 → h = vi2 / 2g
(⊘)
S
mg
F
- Lmolla > 0 = ½ kx2
- Lmolla < 0 = ½ kx2
x > 0
5
5
O →
1
2
0
Compressione Stiramento
3)
m
Ux=0K
UmaxK
x=0, x2=x
Ux(x1)=0
U(x2)= 1/2 kx
U(h)= mgh
U1+K1=O2+K2 1/2 kx o=0
x K/m
Ug1+U1+K1=O2+Ug2+K2
U(x)= 1/2 Kx2
U(y)=mgy
dV=-dL=-F(x)dx
dU/dx O
dU2/dx
F(x)=-dU/dx
Equilibrio
X2
Equilibrio instabile