Calculating Work Required to Stretch a Spring by Additional Distance

Click For Summary
SUMMARY

The discussion centers on calculating the work required to stretch a spring by an additional distance using Hooke's Law. The initial work of 6 J to stretch the spring by 2.0 cm leads to the conclusion that the work required to stretch it an additional 4 cm is 48 J. The calculation utilizes the relationship between work and displacement, confirming that the additional work can be derived without explicitly calculating the spring constant. The key formula applied is W = 1/2 k x^2, and the final result is validated through a proportional reasoning approach.

PREREQUISITES
  • Understanding of Hooke's Law and spring mechanics
  • Familiarity with the work-energy principle in physics
  • Ability to manipulate equations involving variables and constants
  • Knowledge of unit conversions, specifically between centimeters and meters
NEXT STEPS
  • Study the derivation and applications of Hooke's Law in various contexts
  • Learn about energy conservation principles in mechanical systems
  • Explore advanced spring dynamics, including damped and driven oscillations
  • Investigate the relationship between force, work, and potential energy in elastic systems
USEFUL FOR

Students in physics, mechanical engineers, and anyone interested in understanding the principles of elasticity and energy in spring systems.

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?
 
Physics news on Phys.org
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}##.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 17 ·
Replies
17
Views
1K
Replies
2
Views
7K
Replies
29
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
17
Views
2K
Replies
6
Views
981
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K