A spring（k=400N/m)is hung vertically.a) if a 5kg mass is attached to the end if the spring and gently lowered to rest position,what will be the stretch of the spring?b) if the 5kg mass is attached and simply dropped,what will be the maximum velocit

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A spring（k=400N/m)is hung vertically.a) if a 5kg mass is attached to the end if the spring and gently lowered to rest position,what will be the stretch of the spring?b) if the 5kg mass is attached and simply dropped,what will be the maximum velocit

A spring（k=400N/m)is hung vertically.a) if a 5kg mass is attached to the end if the spring and gently lowered to rest position,what will be the stretch of the spring?b) if the 5kg mass is attached and simply dropped,what will be the maximum velocit
A spring（k=400N/m)is hung vertically.a) if a 5kg mass is attached to the end if the spring and gently lowered to rest position,what will be the stretch of the spring?b) if the 5kg mass is attached and simply dropped,what will be the maximum velocity and fhe maximum stretch?

A spring（k=400N/m)is hung vertically.a) if a 5kg mass is attached to the end if the spring and gently lowered to rest position,what will be the stretch of the spring?b) if the 5kg mass is attached and simply dropped,what will be the maximum velocit
第一问：
根据胡克定律 F=x*k F=5*9.81=49.05N
X=F/k=49.05/400=0.123m 因此 第一问的形变是0.123米
第二问：
显然由第一问得出 当形变是0.123米时 弹力和重力首次持平 此时正向加速度为0 因此速度也为最大值
由能量守恒定律可知 1/2mv0²+mgh=1/2mvt²+1/2kx²
上式中 mgh=49.05*0.123=6.03315J
1/2kx²=0.5*400*0.123²=3.0258J
所以可知 1/2mvt²=6.00315-3.0258=3.00735J
vt=1.097m/s
最大形变 当末速度首次为0时 为最大形变
由能量守恒可知
mgh=1/2kh² 解得h=2mg/k=0.24525m

此题用能量守恒的方法 不考虑加速度的中间变化过程
关键是分析什么时刻速度最大 什么时刻形变最大
希望能帮到你

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