<p>a woman weighing 600N steps on a bathroo containing a stiff spring.In equilibrium the spring is compressed 1 cm under her weight.Find the total work done on it during the compression.(force constant=6 x 10^4)
I know the problem can be done by 1/2 <em>k</em>x^2 but i do not understand why can't i just use weigh W=Fs=mgs=600*0.01?The force required to push the string is not constant.but i am applying a constant force,thank you guys.</p>
<p>The problem asks you for the total work done on the spring. So what matters is not the force you apply, but the force on the spring.</p>
<p>W = 600 N
k = 60,000 N/m
x = 0.01 m</p>
<p>Work = U - Uo = .5kx^2 - 0 = .5(60,000)(.01)^2 = 3 J</p>
<p>You can’t solve it the way you want to because you’re forgetting about the spring force pushing back on the person. It takes work to overcome that force, which is done on the spring.</p>
<p>amarkov:but other than the force i applied,what are the other forces acting on it??(because i think its mass is taken to be negligible?or is it the normal force?)</p>
<p>Keasbey:but the force is acting on my but not on the spring,so i think i should not count it to the work?Thanks for explaining guys</p>
<p>It’s the normal force. Whatever the spring is mounted on is exerting a force on it (obviously, or it would just fall through the floor). And that isn’t constant; it depends on how much force the spring is exerting on the woman, which depends on just how far the spring has contracted.</p>
<p>You have to exert a force on the spring to compress it, and in the act of compressing the spring, you are doing work. Work is change in energy, and the change in energy is the spring’s potential.</p>
<p>Thanks guys i got it=)</p>