extension of a spring

procrastin

hey all, this is my first post, sorry for not introducing myself formally, but i am on a tight schedule, getting ready for my retakes.

my question is;

a 5kg monkey, initially at rest, starts climbing up the weightless rope at 0.2g. the top end of the spring is connected to a spring with k=10N/cm. what will be the extension of the spring? what will it be without the monkey?

i have done the first part of the question as follows;

0.2g + g = 1.2g
F= ma
F=5kg x 1.2g = 60N

F = -kx
x= F/-k
x = 60N/-10N/cm
x = -6cm=6cm

(please correct me if i'm wrong)

the part of the question i'm completely stuck on is the second bit; 'what will it be without the monkey?'

if anybody could point me out in the right direction with this part, it would be great. thank you

Doc Al

That's an odd question. Are you sure it didn't say 'what would it be without the monkey climbing?'.

procrastin

yes i am sure. it may be a typo but i'm not risking not doing it without another opinion

Doc Al

Well, without the monkey there would be no extension at all. (We're assuming that the spring is massless, of course.)

procrastin

ok i'll solve the question with the assumption that there is a typo. do u know if the previous part of he question i've answered is correct?

Doc Al

Yes, it's fine. (But I recommend thinking in terms of forces acting on the monkey, instead of directly in terms of acceleration.)

procrastin

do u mean like this;

F1=ma=(5kg)(0.2g)
F2=mg=(5kg)(g)
Ftotal=ma+mg
x=Ftotal/-k

?

Doc Al

I look at it like this. There are two forces acting on the monkey: the tension pulls up, the weight pulls down:

∑F = ma
T - mg = ma

Thus T = ma + mg.

And the tension is the force that extends the spring, so from Hooke's law: T = kx, or x = T/k.

Same thing, expressed a bit cleaner.

procrastin

i understand, thank you for your help <3