题目

(16分)某同学通过实验测定一段坡路的倾角,该坡路通过小圆弧过渡与水平路面相接。他将一小物体从坡路上某点由静止释放,物体到达坡底后又在水平路面上滑行一段距离后停下,该同学测得物体在坡路及水平路面上滑行的距离分别为s1=4m和s2=1.6m,所用总时间t=2.8s。该坡路及水平地面与物体间动摩擦因数相同,g取10m/s2。请你根据这些数据,计算该坡路的倾角。 答案:解析:设坡路倾角为,物体在坡路上下滑加速度a,根据牛顿定律得                                                          (1)(4分)物体滑到斜坡底端时速度为,则:                                                                                        (2)(2分)                                                                                       (3)(2分)                                                                                      (4)(2分)设物体在水平路面上加速度为1                                                                                   (5)(2分)                                                                                      (6)(2分)代入数据可解得:(7)(2分)
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