Gọi t là thời gian đi hết quãng đường

\(\left\{{}\begin{matrix}v_0t+\frac{1}{2}at^2-v_0\left(t-1\right)-\frac{1}{2}a\left(t-1\right)^2=1\\a=\frac{v-v_0}{t}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2}at^2+v_0-\frac{1}{2}at^2+at-\frac{1}{2}a=1\\v_0+at=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v_0+at-\frac{1}{2}a=1\\v_0+at=0\end{matrix}\right.\)

\(\Leftrightarrow-\frac{1}{2}a=1\Leftrightarrow a=-2\left(m/s^2\right)\)

=> \(F_h=\left|m.a\right|=\left|500.\left(-2\right)\right|=1000\left(N\right)\)

la:46;48;50

k nha

\(5+7+18=30\)

Vật trượt đều=> a= 0

Xét F hướng lên

Theo định luật II Niu-tơn:

\(\overrightarrow{F_{ms}}+\overrightarrow{P}+\overrightarrow{N}+\overrightarrow{F}=\overrightarrow{0}\)

\(\Rightarrow\left\{{}\begin{matrix}Ox:-P.\sin\alpha-F_{ms}+F=0\\Oy:-P.\cos\alpha+N=0\end{matrix}\right.\)

\(\Leftrightarrow-mg\sin\alpha-\mu mg\cos\alpha+F=0\)

\(\Leftrightarrow-0,1.10.\sin\alpha-\sqrt{3}.0,1.10.\cos\alpha+10=0\)

\(\Leftrightarrow\sin\alpha+\sqrt{3}\cos\alpha=10\) (1)

Ta sẽ lập 1 pt nx:

Khi lực F hướng xuống:

\(\left\{{}\begin{matrix}Ox:P.\sin\alpha+F-F_{ms}=0\\Oy:-P.\cos\alpha+N=0\end{matrix}\right.\)

\(\Leftrightarrow mg\sin\alpha+10-\mu mg\cos\alpha=0\)

\(\Leftrightarrow0,1.10.\sin\alpha-\sqrt{3}.0,1.10.\cos\alpha=-10\)

\(\Leftrightarrow\sin\alpha-\sqrt{3}\cos\alpha=-10\) (2)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sin\alpha+\sqrt{3}\cos\alpha=10\\\sin\alpha-\sqrt{3}\cos\alpha=-10\end{matrix}\right.\)

Bạn tự giải nốt để tìm alpha

262260437

k cho mk nha 

11-8=3 (cái kẹo )

1 + 2 + 3 + 4 = 10