对小滑块受力分析，如图所示

竖直方向，有
$N_{1} + E q \sin θ = m g$
解得
$N_{1} = m g - E q \sin θ = 0.5 \times 10 N - 500 \times 0.01 \times 0.6 N = 2 N$

$f_{1} = μ_{1} N_{1} = 0.5 \times 2 N = 1 N$

水平方向根据牛顿第二定律
$q E \cos 37 ° - f_{1} = m a_{1}$

联立解得

$a_{1} = \frac{q E \cos 37 ° - f_{1}}{m} = \frac{0.01 \times 500 \times 0.8 - 1}{0.5} m / s^{2} = 6 {m/s}^{2}$

小滑块位移

$x_{1} = \frac{1}{2} a_{1} t^{2} = \frac{1}{2} \times 6 \times 1^{2} m = 3 m$

对长木板受力分析，如图所示

竖直方向
$N_{2} = N' + M g = m g - E q \sin 37 ° + M g = 0.5 \times 10 N - 500 \times 0.01 \times \sin 37 ° + 1 \times 10 N = 12 N$
水平方向，根据牛顿第二定律，有

$f_{2} = μ_{2} N_{2} = 0.05 \times 12 N = 0.6 N$

${f^{'}}_{1} - f_{2} = M a_{2}$

其中

$f_{1}' = {f^{}}_{1} = 1 N$

联立解得

$a_{2} = \frac{f'_{1} - f_{2}}{M} = \frac{1 - 0.6}{1} {m/s}^{2} = 0.4 {m/s}^{2}$

长木板位移

$x_{2} = \frac{1}{2} a_{2} t^{2} = \frac{1}{2} \times 0.4 \times 1^{2} m = 0.2 m$

长木板长度

$L = x_{1} - x_{2} = 3 m - 0.2 m = 2.8 m$

湖北省云学部分重点高中2024-2025学年高二上学期9月月考物理试题