Hooke's Law and springs

  • Thread starter mateomy
  • Start date
  • #1
307
0
A 2.0 kg mass and a 3.0 kg mass are on a horizontal frictionless surface, connected by a massless spring with spring constant k=140N/m. A 15 N force is applied to the larger mass, as a shown (see picture). How much does the spring stretch from its equilibrium length?


hookesExample.jpg




I've solved this problem -the answer is 4.28cm- but I had to do it using the force on the smaller of the masses (2 kg). My question is why I have to take it from that mass? Why couldn't you just find it from the 3 kg mass, by just taking the initial 15 N force as proportional to the Force of the spring? If you do this the book says that answer is wrong. The physical interpretation of WHY we have to to find it from the 2kg force is eluding me.
 

Answers and Replies

  • #2
Chronos
Science Advisor
Gold Member
11,408
738
The direction of force makes all the difference.
 

Related Threads on Hooke's Law and springs

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
1
Views
970
  • Last Post
Replies
15
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
9K
Top