Question

如圖所示，質(zhì)量為m=2kg的小球系在輕質(zhì)彈簧的一端，另一端固定在懸點(diǎn)O處，將彈簧拉至水平位置A處，且彈簧處于自然狀態(tài)，彈簧的原長0A=0.3m；然后小球由靜止釋放，小球到達(dá)距O點(diǎn)下方h=0.5m處的B點(diǎn)時(shí)速度為V_B=2m/s，求                                                                                                                                              

（1）小球從A運(yùn)動(dòng)到B的過程中彈簧的彈力做的功和此時(shí)彈簧的彈性勢能．                                

（2）求該彈簧的勁度系數(shù)．                                                                                          

                                                                                                           

                                                                                                                                      

Answer 1

（1）用動(dòng)能定理：設(shè)彈簧做功 W，則

mgh+W=mV_B²﹣0

帶入后，可得：W=﹣6J

也就是說彈簧做功﹣6J，

彈簧的彈力做的功量度彈簧彈性勢能的變化，所以此時(shí)彈簧的彈性勢能為6J．

（2）在B 點(diǎn)，彈簧形變量x=0.5﹣0.3=0.2m

根據(jù)E_P=

解得K=300N/m

答：（1）小球從A運(yùn)動(dòng)到B的過程中彈簧的彈力做的功是﹣6J 和此時(shí)彈簧的彈性勢能是6J．

（2）求該彈簧的勁度系數(shù)是300N/m．