m = 5 kg
v = 3,6 km/h = 3,6*1000/3600 = 3600/3600 = 1 m/s
Δt = 10s
[tex]\mu[/tex] = ct
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a) Ec(A) = ?
b) LFf = ?
c) Ff = ?
d) d = ?
e) Ec(B) = ?
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a)
[tex]\displaystyle\\\rm\\E_{c(A)}= \frac{mv^{2}}{2} = \frac{m}{2} = \frac{5}{2} = 2,5 \ Jouli[/tex]
d)
- se aplica legea vitezei in miscarea rectilinie uniform variata (MRUV)
[tex]\displaystyle\\\rm\\v = v_0 + at[/tex]
[tex]\displaystyle\\\rm\\v = 0 \ iar \ v_0 = v_A \implies v_A = -at[/tex]
[tex]\displaystyle\\\rm\\a = - \frac{v_A}{t} = -\frac{1}{10} = -0,1 \ m/s^2[/tex]
(acceleratia este negativa deoarece corpul incetineste)
- se aplica legea miscarii in MRUV
[tex]\displaystyle\\\rm\\d = d_0 + v_0 \cdot (t - t_0) + \frac{a(t-t_0)^2}{2}[/tex]
- la noi, [tex]t_0 = 0[/tex], [tex]d_0 = 0[/tex] si [tex]v_0 = v_A = 1 \ m/s[/tex]
- formula devine:
[tex]\displaystyle\\\rm\\d = v_A \cdot t + \frac{at^{2}}{2}[/tex]
[tex]\displaystyle\\\rm\\d = 1 \cdot 10 + \frac{-0,1 \cdot 10^2}{2}[/tex]
[tex]\displaystyle\\\rm\\d = 10 - \frac{100 \cdot 0,1}{2}[/tex]
[tex]\displaystyle\\\rm\\d = 10 - \frac{10}{2}[/tex]
d = 10 - 5
d = 5 m
- deci corpul se va opri dupa 5 metri
b si c)
- se aplica Principiul II pe axa Ox
-Ff = m*a
Ff = -m*a
Ff = 0,1 * 5
Ff = 0,5 N
[tex]\displaystyle\\\rm\\L_{Ff} = -Ff \cdot d[/tex]
[tex]\displaystyle\\\rm\\L_{Ff} = -0,5 \cdot 5[/tex]
[tex]\displaystyle\\\rm\\L_{Ff} = -2,5 \ Jouli[/tex]
e)
- pentru t' = 3s, se aplica din nou legea vitezei
[tex]\displaystyle\\\rm\\v = v_A + a \cdot (t' - t_0)[/tex]
[tex]\displaystyle\\\rm\\v = 1 - 0,1 \cdot (3 - 0)[/tex]
[tex]\displaystyle\\\rm\\v = 1 - 0,1 \cdot 3[/tex]
[tex]\displaystyle\\\rm\\v = 1 - 0,3 = 0,7 \ m/s[/tex]
- la t' = 3s, viteza corpului este de 0,7 m/s
[tex]\displaystyle\\\rm\\E_c = \frac{m \cdot v^2}{2}[/tex]
[tex]\displaystyle\\\rm\\E_c=\frac{5 \cdot 0,7^2}{2}[/tex]
[tex]\displaystyle\\\rm\\E_c = \frac{5,49}{2}[/tex]
Ec = 2,745 Jouli