高一物理关于弹簧的问题

2024-12-28 13:52:24
推荐回答(3个)
回答1:

解析:
问题一:前半段过程(到平衡点之前)
物体刚放在弹簧上时候,受力竖直向上重力,开始向下运动时受到F弹=kx,所以F合=mg-F弹=ma,随着压缩量增大F弹变大,a变小,但是初始速度向下a也是向下,a与V同向物体做的是加速度减小的加速运动。当F弹增大到等于mg之时(平衡点)速度达到最大此时加速度为零。

问题二:后半段过程(平衡点到最低点)过了平衡点,F弹再增大比mg要大了,此时合外力为F合=F弹-mg=ma,压缩量还在变大F弹也就在变大,a变大,但此时a与V是反向的,所以物体做加速度增大的减速运动,到达最低时候,速度减为0了。

问题三:综合问题二和问题一的讲解。前半段a与速度都是向下的,a变小的加速运动。后半段a是向上v是向下,a变大的减速运动。

回答2:

1.B球受力:剪断细线前,重力向下mg,细线向上拉力4mg,所以,弹簧向下弹力4mg-mg=3mg.剪断细线,4mg消失,重力和弹力还存在,所以,合力向下为3mg+mg=4mg,加速度为4mg/m=4g.
2.A球受力:剪断细线前,重力向下mg,细线向下拉力4mg,下面弹簧向上弹力3mg,所以,上面弹簧向上拉力T=mg+4mg-3mg=2mg.剪断细线,4mg消失,重力和弹力还存在,所以,合力向上为3mg+2mg-mg=4mg,加速度为4mg/m=4g.

回答3:

1,是
2,不是,速度为零
3,落到弹簧前,速度增大,加速度不变,
落到弹簧后速度变小,加速度向上

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