What work is required to stretch a spring of spring constant 40N/m from x =0.20m to 0.25m?(the unstretched position is at x =0) a) 0.45J, b)0.80J, c) 1.3J, d) 0.050J, e) 0.90J Please , help

Question

loser66 · Answer

My work is k =40N/m , x1=0.20m so, W1=12kx21 and x2=0.45m W2=12kx22 △W=W2−W1=12k(x22−x21)=12∗40∗(0.452−0.202)=2.8J However, it's wrong, I don't know why they choose a) 0.45J

loser66 · Answer

some mess there. hehe. sorry, $W_1= 1/2 k x_1^2$ and $W_2 = 1/2 kx_2^2$

fifciol · Answer

$$W=\frac{ 1 }{ 2 }k(x_2^2-x_1^2)=20(0.0625-0.04)=0.45J$$

loser66 · Answer

$x_2^2= 0.45^2 = 0.2025$

fifciol · Answer

0.25^2

loser66 · Answer

ok, I know what I am wrong, hehehe.. My bad English.

loser66 · Answer

Thanks a ton. It moves to 0.25 m not move 0.25m more