消防员进行滑杆下楼训练，一质量为60kg的消防员从脚离地20m的杆

问题问答题

消防员进行滑杆下楼训练，一质量为60kg的消防员从脚离地20m的杆上由静止开始下滑，为确保安全，中间过程的最大速度不得超过15m/s，着地时的速度不得超过4m/s，消防员和杆间能获得的最大阻力为900N，g=10m/s²．求：

（1）消防员减速下滑过程的最大加速度多大？

（2）在确保安全的情况下消防员下楼的最短时间是多少？

答案

（1）减速下楼过程的加速度大小为a

f-mg=ma

当阻力取最大值时，加速度最大，则

f-mg

900-60×10

=5m/s2

（2）消防员要使下楼时间最短，则他应先自由下落，再以最大加速度减速．如果消防员能达到最大速度v=15m/s，则其下楼过程的位移至少为s

v2-vmin2

152

152-42

=31.15＞20m

由此可以判断，消防员下楼过程不能达到15m/s，设实际最大速度为v_m

vm2

vm2-vmin2

带入数据解得：

v_m=12m/s

自由下落过程时间为t₁，减速下楼过程时间为t₂

t1=

=1.2s

t2=

vm-vmin

12-4

=1.6s

总时间t=t₁+t₂=1.2+1.6=2.8（s）

答：（1）消防员减速下滑过程的最大加速度为5m/s²；

（2）在确保安全的情况下消防员下楼的最短时间是2.8s．

完形填空。
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