F (t) a(t) v(t) x(t) ??

A = F · d

(1)

A F d [A] = [F]·[d] = =

A = F ·d = 4·5 = 20 1cal = 4, 186 et 1 = 3, 6e6

?? ??

P = \frac{A}{t}

(2)

P A

[P] = \frac{[A]}{[t]} = \frac{J}{s} =

P = \frac{A}{t} = \frac{20}{60} = 0, 33

1Ch = 736

?? A = P ·t

[A] = [P]·[t] =

m h A = F ·d = P ·h = mg·h

h h mg·

h

E_{pot} = mg · h

(3) [E_{pot}] = [A] =

h m h

v^{2} = v_{o}^{2}+2·a·d ⇒ a = \frac{v^{2} − 0^{2}}{2 · d}

·d = ma· d = m· \frac{v^{2}}{2·d}·

d = \frac{1}{2}· m· v^{2} m v

E_{cin} = \frac{1}{2} · m · v^{2}

(4) [E_{cin}] = [A] =

m ??

ΔE_{cin} = ∑ A_{F^{ext}}

(5)

?? ?? cin_{p}ot_{c}irc.eps

h E_{pot} = m·g·h

E_{cin} = 0

h ↘ ⇒ E_{pot} = m·g·h ↘

v ↗ ⇒ E_{cin} = \frac{1}{2}·m·v^{2} ↗

h = 0 ⇒ E_{pot} = 0 ; E_{cin} = \frac{1}{2}·m·v^{2}

conservation de l’nergie

E_{pot} = mg·h = \frac{1}{2}·m·v^{2} = E_{cin}

·h = \frac{1}{2}· m· v^{2} ⇒ v = \sqrt{2 · g · h}

principe de conservation de l’nergie nergie mcanique