问题 计算题

(16分)

如图所示,在高1.25m的水平桌面上,一质量为2.0kg的物块在10N的水平拉力作用下, 在A处由静止开始向桌面边缘B运动,2s末撤去水平拉力。物块运动到桌面B端后飞出落在水平地面上。已知物块与桌面之间的动摩擦因数μ=0.3,AB之间的距离为6m,不计空气阻力,g=10m/s2。求:

(1)撤去水平拉力前物块加速度的大小;

(2)物块离开桌面边缘B点时速度的大小;

(3)物块落地点距桌面边缘B点的水平距离。

答案

(1)

(2)

(3)

(1)(4分)物块在水平高台上做匀速直线运动,设加速度为a1,由牛顿第二定律

                                                                            …………2分

                                                                               …………2分

(2)(8分)设物块运动2s的位移S1、2s未的速度v1

                                                                       …………2分

                                                                       …………2分

撤去水平拉力后物块加速度为a2,物块离开桌面上时速度为v2

                                                                         …………2分

                                                                           …………1分

                                                                                …………1分

(3)(4分)物块离开桌面后做平抛运动,设落地点物块距高台边缘水平距离为L

                                                                                    …………1分

                                                                                      …………1分

                                                                                        …………1分

                                                                                      …………1分

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