在升降机中的水平底板上放一质量为60 kg的物体,如图4-6-10所示为升

在升降机中的水平底板上放一质量为60 kg的物体,如图4-6-10所示为升降机下降过程中的v-t图象,试通过计算作出底板对物体的支持力F随时间t的变化图象.(g取10 m/s2

                                                                 图4-6-10

答案

解析:速度图线显示了整个过程中的速度大小和变化规律,所以本题是已知运动求力的问题.

    运动过程分三个阶段:

    第一阶段,初速度为零的向下匀速运动,由图象得a1= m/s2=2 m/s2

    由牛顿第二定律得mg-F1=ma1,F1=m(g-a1)=60×(10-2) N=480 N

    第二阶段,匀速向下运动,物体处于平衡状态,由牛顿第二定律得mg-F2=0,

    即F2=mg=600 N

    第三阶段,向下匀减速运动,由15—25 s间的速度图象可知a3= m/s2=-1 m/s2

    由牛顿第二定律得mg-F3=ma3

    F3=m(g-a3)=60×[10-(-1)] N=660 N

    由以上分析作出0—25 s时间内水平地板对物体的支持力F随时间变化的图象,如图所示.

答案:见解析


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