How Far Does the Block Compress the Spring?

Click For Summary
SUMMARY

The discussion focuses on a physics problem involving a 1.40 kg block sliding down a 20° inclined plane towards a spring with a spring constant of 400 N/m. The block is initially 2.30 meters away from the spring and experiences no friction. The key equations used include the conservation of energy, specifically the potential energy (PE) equations PE = 1/2 kx² and PE = mgh. The maximum compression of the spring and the velocity of the block when it loses contact with the spring are the primary objectives of the problem.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with spring mechanics and Hooke's Law
  • Knowledge of basic trigonometry related to inclined planes
  • Ability to solve equations involving potential energy and kinetic energy
NEXT STEPS
  • Study the derivation and application of conservation of energy in mechanical systems
  • Learn about Hooke's Law and its implications for spring compression
  • Explore the relationship between potential energy and kinetic energy in motion on inclined planes
  • Practice solving similar problems involving inclined planes and springs
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of blocks on inclined planes and the mechanics of springs.

MUFfL3r
Messages
1
Reaction score
0

Homework Statement



A block with a mass of 1.40 kg is free to slide down a plane inclined at an angle of 20° above horizontal (the block will slide down and to the left). Before it is released from rest, it is a distance of d = 2.30 meters away from the end of a spring with spring constant 400 N/m. Assume there is no friction. g = 9.80 m/s²

mass = 1.40 kg
ϴ = 20°
d = 2.30 m
k = 400 N/m
g = 9.80 m/s²a) When the the compression of the spring is a maximum, what is the distance that the spring is compressed past equilibrium point?

b) When the block is on the way back up the ramp, how fast is it going at the moment it loses contact with the spring?

Homework Equations



Conservation of energy:
PE = 1/2kx²
PE = mgh

The Attempt at a Solution



a) I worked it out to mgd(sinϴ) = -mgd(sinϴ) + ½kx² is that right?

b)Well I know I need to find velocity. But I'm having a hard time trying to figure out how I would relate the equations together. Any helpful hints?
 
Last edited:
Physics news on Phys.org
Draw a picture and show both the initial and the final positions of the block, when the spring is compressed and the block stopped.

ehild
 
A bit of a hint for the velocity: If there is no friction, then energy is going to be in either the block or the spring at any given time.
 
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 12 ·
Replies
12
Views
3K
Replies
17
Views
2K
  • · Replies 27 ·
Replies
27
Views
10K
Replies
8
Views
6K
  • · Replies 18 ·
Replies
18
Views
5K
Replies
3
Views
1K
Replies
24
Views
4K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K