Stretching a Spring

  • #1
RGBolton95
Hey everyone,

I'm really hoping somebody will be able to help me with this problem. I've searched all through my textbook, notes, and the Internet, but I keep getting the wrong answer. Here's the question:

If it requires 6 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be required to stretch it an additional 4 cm?

I start by finding k, the spring constant. I do this by using the equation W = 1/2kx^2. I plug in the known values of work and x in the equation, which gives me 30,000 Nm (I changed 2 cm to 0.02 m). Now that I know k, I find the total work to pull the spring down 0.06 m (the original 0.02 m plus the additional 0.04 m). I use the same W = 1/2kx^2 equation, with W being the unknown, k = 30,000 Nm, and x = 0.06. This gave me a W = 54 J. From there, I know that to find the additional work needed to pull down the spring from 0.02 m to 0.06 m I should just subtract 6 J from 54 J, which gives me a final answer of 48 J. However, I keep getting told that this is the wrong answer. Any idea where I'm making my mistake?
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,011
6,806
Your result is correct.

You should double check the units of your spring constant though.

Note that you really do not need to compute the spring constant, it is sufficient to note that
$$
2W_2 = k x_2^2 = k(3x_1)^2 = 9kx_1^2 = 18W_1
$$
and therefore ##W_2 = 9W_1## and ##W_2-W_1 = 8W_1 = 8\cdot 6\ \mathrm{J} = 48\ \mathrm{J}##.
 

Related Threads on Stretching a Spring

Replies
2
Views
3K
  • Last Post
Replies
6
Views
9K
Replies
3
Views
925
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
870
  • Last Post
Replies
3
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
2K
Top