Question

Un corp de masa m= 50 kg
.. problema 2

Answer 1

$m=50 \ kg \\ h=4 \ m \\ \alpha =30^o \\ \mu=\frac{\sqrt{3}}{4} \\ \vec{g}=10 \ m/s^2 \\ \vec{a}_{\downarrow}=\vec{g}(\sin \alpha-\mu \sin \alpha) \\ \Rightarrow \vec{a}_{\downarrow}=10 \ m/s^2(\sin 30^o-\frac{\sqrt3}{4} \cos 30^o) \\ \Rightarrow \vec{a}_{\downarrow}=10 \ m/s^2(\frac{1}{2} -\frac{\sqrt3}{4} \cdot \frac{\sqrt3}{2}) \\ \Rightarrow \vec{a}_{\downarrow}=10 \ m/s^2(\frac{4}{8}-\frac{3}{8})=10 \ m/s^2 \cdot \frac{1}{8}=\frac{10 \ m/s^2}{8} \\ \Rightarrow \boxed{\vec{a}_{\downarrow}=1,25 \ m/s^2};$

$E_c=\frac{mv^2}{2}; \\ \vec{a}=\frac{\vec{v}}{t} \ \Rightarrow \vec{v}=\vec{a}t=\frac{d}{t}=\frac{l}{t} \\ l=2h=2 \cdot 4 \ m=8 \ m; \\ \Rightarrow l=\vec{a}t^2 \ \ \Rightarrow t=\sqrt{\frac{l}{\vec{a}}}=\sqrt{\frac{8 \ m}{1,25 \ m/s^2}} =\sqrt{6,4} \ s\\ \Rightarrow t=2,53 \ s; \\ \Rightarrow \vec{v}=1,25 \ m/s^2 \cdot 2,53 \ s=3,16 \ m/s=\sqrt{10} \ m/s; \\ \Rightarrow E_c=\frac{50 \ kg \cdot (\sqrt{10} \ m/s)^2}{2}=\frac{500 \ J}{2} \\ \Rightarrow \boxed{E_c=250 \ J}$