Observam bine din punctul b), ca miscarea nu e deloc uniforma, deci vom lua in consideratie ca in toate punctele e aceeasi putere.
[tex]Se~da:\\ \\
m=200t=200000kg\\ \\
P=400kW=400000W\\ \\
\mu=0,01\\ \\
a)v_m=?\frac ms\\ \\
b)v_1=1\frac ms;~v_2=10\frac ms\\ \\
a_1=?\frac m{s^2};~a_2=?\frac m{s^2}\\ \\
c)t=2s;~L=?J\\ \\ \\[/tex]
[tex]Formule:\\ \\
a)P=F\times v_m\\ \\
v_m=\frac PF\\ \\ \\
Daca~e~atinsa~viteza~maxima,~deci~miscarea~devine~uniforma.\\ \\
Conform~legii~lui~Newton,~daca~viteza~e~uniforma:\\ \\
F=F_{fr}\\ \\
F=\mu \times N\\ \\
F=\mu\times m\times g\\ \\ \\
v_m=\frac P{\mu \times m \times g}\\ \\ \\[/tex]
[tex]b)Observam~o~miscare~accelerata,~deci~cf.~legii~lui~Newton:\\ \\
F-F_{fr}=m\times a\\ \\
a=\frac{F-\mu\times m\times g}m\\ \\
a=\frac Fm-\mu\times g\\ \\
P=F\times v\\ \\
F=\frac Pv,~deci:\\ \\ \\
a_1=\frac P{v_1\times m}-\mu\times g\\ \\
a_2=\frac P{v_2\times m}-\mu\times g\\ \\ \\
c)L=P\times t\\ \\ \\
[/tex]
[tex]Calcule:\\ \\
a)v_m=\frac {400000}{0,01 \times 200000 \times 10}=20\frac ms\\ \\
b)a_1=\frac {400000}{1\times 200000}-0,01\times 10=1,9\frac m{s^2}\\ \\
a_2=\frac {400000}{10\times 200000}-0,01\times 10=0,1\frac m{s^2}\\ \\
c)L=400000\times 2=800000J[/tex]
