Calculating the Extension of a Spring with and without a Hanging Mass

[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]

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

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

 

[procrastin]

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

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]

do u know if the previous part of he question I've answered is correct?

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]

do u mean like this;

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

?

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