突然腦閉塞唔記得左點做-_-

A spring of force constant 140Nm-1 is suspended vertically with its top end rigidly clamped . A mass of 0.14kg is attached to the lower end of the unstrechtd spring and released .

Calculate the extension of the spring when the mass reaches the lowest position .

先謝

唔知係咪咁做

F = kx
mg = kx
x = mg/k
= 0.14x10/140
= 0.01m

坤仔 :唔知係咪咁做

F = kx
mg = kx
x = mg/k
= 0.14x10/140
= 0.01m

mg=kx唔係equi. position先係架咩...
(拉力F balanced左Weight)



ps:答案係0.02m

mgx=1/2kx^2
1.4x=70x^2
x=0.02

成條數應該係計緊個potential difference convert去extension of the spring
但就唔係好明關SHM咩事...

Snoopy@J&H :mgx=1/2kx^2
1.4x=70x^2
x=0.02

成條數應該係計緊個potential difference convert去extension of the spring
但就唔係好明關SHM咩事...

GPE in equi. > EPE in lowest point ？
如果係咁明晒

ps:條數本身放左係SHM果課,
而SHM我都記得有用教過energy change

anyway thx