Velocity Of An Object Accelerated By A Spring

  • Thread starter Thread starter fcv33
  • Start date Start date
  • Tags Tags
    Spring Velocity
Click For Summary
SUMMARY

The velocity of an object released from a compressed spring can be calculated using the formula v = sqrt(k/m) * x, where k is the spring constant, m is the mass of the object, and x is the initial displacement of the spring. The discussion confirms that the kinetic energy of the object at the moment it leaves the spring equals the potential energy stored in the spring, represented by the equation E = 0.5kx^2. The integration approach also leads to the same result, reinforcing the validity of the derived formula.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Basic principles of energy conservation
  • Knowledge of calculus for integration
  • Familiarity with kinetic and potential energy equations
NEXT STEPS
  • Study the derivation of Hooke's Law and its applications in mechanics
  • Explore energy conservation principles in mechanical systems
  • Learn about integration techniques in physics for motion analysis
  • Investigate real-world applications of spring dynamics in engineering
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the dynamics of spring systems and energy conservation principles.

fcv33
Messages
1
Reaction score
0

Homework Statement


I'm pulling my hair out over this problem. A spring with a force constant of k is compressed x meters. An object of mass m is placed against the spring. The spring is then released. At what velocity is the object moving at the instant that is ceases to touch the spring?


Homework Equations


E=0.5kx^2

The Attempt at a Solution


I'm really unsure. All of my solutions don't seem realistic.
 
Physics news on Phys.org
Using Energy:
E=0.5kx^2=Kinetic Energy=0.5mv^2.

solve for v=((k/m)^1/2)*x
where x is initial displacement of spring.

Using F
F=-kx x is negative therefore
F=kx=ma

a=dv/dx*dx/dt = dv/dx*v
vdv=(kxdx)/m integrate
0.5v^2=0.5k/m x^2
v=((k/m)^1/2)*x
 

Similar threads

Replies
29
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 14 ·
Replies
14
Views
3K