How Fast Does the Block Travel After Leaving the Spring?

Click For Summary
SUMMARY

The discussion focuses on the mechanics of a 2.00-kg block released from a compressed spring with a force constant of 400 N/m and a compression of 0.220 m. The block moves along a frictionless surface and ascends a frictionless incline at a 37-degree angle. Energy conservation principles are applied, where the potential energy stored in the spring is converted into kinetic energy as the block moves. The kinetic energy is then transformed into gravitational potential energy as the block travels up the incline.

PREREQUISITES
  • Understanding of energy conservation principles in physics
  • Familiarity with potential energy and kinetic energy equations
  • Knowledge of spring mechanics, specifically Hooke's Law
  • Basic trigonometry for calculating forces on an incline
NEXT STEPS
  • Study the conservation of mechanical energy in closed systems
  • Learn about Hooke's Law and its applications in spring mechanics
  • Explore the equations of motion on inclined planes
  • Investigate the relationship between kinetic energy and gravitational potential energy
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding energy transformations in physical systems.

ledhead86
Messages
59
Reaction score
0
A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope of 37 degrees.
Pic
http://community.webshots.com/user/mmaddoxwku"
What is the speed of the block as it slides along the horizontal surface after having left the spring?
 
Last edited by a moderator:
Physics news on Phys.org
If you'd like help, you must show your attempted solution. Hint: Energy is conserved.
 
so I am thinkin in need to use the formula delta_k=-delta_u. Or k_1+U_1=k_2-K_1. So would K_1= (1/2)(2 kg)(0)^2, and U_1=(1/2)(400 N/m)(0.22 m)^2. k_2=(1/2)(2 kg)*V_2^2, and U_2=(1/2)(400 N/m)(0)^2.
 
Try to express what's going on in words:

When compressed the block has _______ energy and no _____ energy, and when released the ____ energy will be converted into ______ energy because now the block will be moving. Since the sufrace is frictionless the block will move at a _____ velocity having a KE equal to the initially stored _____ energy. It will then lose all this ______ energy to become gravitational _____ energy as it goes up the ramp.


After you fill that out correctly, it should be fairly easy to figure out what is equal to what, and so forth to plug in numbers.
 
When compressed the block has potential energy and no kinetic energy, and when released the potential energy will be converted into kinetic energy because now the block will be moving. Since the sufrace is frictionless the block will move at a constant velocity having a KE equal to the initially stored PE energy. It will then lose all this KE energy to become gravitational potential energy as it goes up the ramp.

So I had it setup correctly. Now how do I find how far the block travels up the incline before starting to slide back down ?
 
ledhead86 said:
So I had it setup correctly. Now how do I find how far the block travels up the incline before starting to slide back down ?
By finding the point where all the energy is transformed to gravitational PE.
 
OK. So I know -delta_U=F* D * sin(theta)
So. 9.68=F*D*sin(37). What is F. Is that 400N, the spring constant?
 
Never mind, I figured it out.
 
Did the fill inthe blanks help?
 

Similar threads

Replies
17
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
14
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 5 ·
Replies
5
Views
10K