如图所示，一带电量为q＝－5×10－3 C，质量为m＝0.1 kg的小物块处于一倾角为θ＝37°的光滑绝缘斜面上，当整个装置处于一水平向左的匀强电场中时，小物块恰处于静止状态．(g取10 m/s2)：
(1) 求电场强度多大？
(2)若从某时刻开始，电场强度减小为原来的
，求物块下滑距离L＝1.5 m时的速度大小
(1) 150N/C(2) 1.35J
题目分析：受力分析：竖直向下的重力，水平向右的电场力，垂直斜面向上的支持力
（1）N*cos37°=mg ①
Eq=N*sin37° ②
联立得 E=(mg*tan37°)/q=150N/C
（2）运用正交分解法
mg*sin37°-0.5*Eq*cos37°=ma
解得 a=0.3g
由动能定理：
mg*L*sin37°+0.5*Eq*L*cos37°=Ek-0
解得 Ek=0.9*mgL=1.35J
点评：本题难度较小，注意重力和电场力做功与路径无关，与初末位置有关
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