Why Do Force and Energy Analyses Yield Different Spring Extensions?

In summary, the conversation discusses the use of force and energy analysis equations to find the maximum extension of a spring attached to a block on an inclined plane. There is a discrepancy between the two methods and the concept of energy conservation is brought up. The potential and kinetic energy of the system are also considered, along with the role of equilibrium and acceleration in the calculations. Finally, the properties and behavior of a spring under compression and expansion are explained.
  • #1
positron
21
0
I have a question about thse following system: block of mass M is attached to a spring is on an inclined plane.
Why do the force and energy analysis equations give different answers? The question is to find the maximum extension of the spring. Here is my thinking so far.


I know if you use energy convervation, you get 1/2*k*x^2 = M*g*sin(theta), so you get x = 2*M*g*sin(theta)/k.
However, why can I not just set the force along the incline equal to the spring force and solve for x that way? If I do that, I get a different answer:
k*x = M*g*sin(theta)
x = M*g*sin(theta)/k and this is different by a factor of 2.
 
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  • #2
What exactly is the situation here? Could you draw a diagram?
What's with the mgsin(theta)? It's a force, not work. You cannot use it in the energy conservation equation.
 
  • #3
The problem is the energy convervation. The potential energy should be
M*g*h were h is the hight difference of the two states (before the mass was released and after).
What about kinetic energy as the weight get to the end? (it has mass and it is moving)
Also, the way that you did the first answer is missing a square route.
If you can find the connection between the h and the x then the ^2 could cancel out!
 
  • #4
In the answer above I assumed that you wanted the equilibrium. If you want the maximum extension the you can't say that the force of gravity and the spring along the plane are equal because as the mass is accelerating as it gets to the bottom (harmonic motion)
 
  • #5
A spring with constant k, and a displacement (x) from static deflection position (dst), also called equilibrium position.

Force: k*dst = W_block = mg
mg - k(x+dst) = ma
ma + kx = 0

ma here gives the force on a spring that expands it by x from equilibrium position, dst. F = -kx, by convention direction of x is negative to the dst, so F = kx

When you compress the spring you are doing mechanical work on it W = (F*dr) and storing energy (KE, PE, or both) in the spring. When the force is removed the spring does work on the block to move it a distance d against external forces. When it is not compressed it has no stored energy and a force is applied to expand the spring from its equilibrium position.
 

FAQ: Why Do Force and Energy Analyses Yield Different Spring Extensions?

What is the relationship between mass and acceleration on an inclined plane?

The relationship between mass and acceleration on an inclined plane is described by the equation F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration. On an inclined plane, the force of gravity is acting on the object, causing it to accelerate down the slope. The steeper the slope, the greater the acceleration will be.

How does the spring constant affect the motion of an object attached to a spring?

The spring constant, represented by the symbol k, determines the stiffness of the spring. The higher the spring constant, the stiffer the spring and the greater the force required to stretch or compress it. This affects the motion of an object attached to a spring by determining the magnitude of the force exerted on the object and thus, its acceleration. A higher spring constant will result in a greater acceleration of the object.

What is the significance of the angle of inclination on an inclined plane?

The angle of inclination on an inclined plane is significant because it determines the steepness of the slope and thus, the acceleration of an object placed on the plane. A steeper incline will result in a greater acceleration, while a shallower incline will result in a slower acceleration. The angle of inclination also affects the normal force acting on the object, which is equal to the component of the force of gravity perpendicular to the plane.

How does mass affect the period of a mass-spring system?

The period of a mass-spring system is the time it takes for the object to complete one full cycle of motion. The mass of the object attached to the spring does not affect the period, as long as the spring constant and amplitude of the motion remain constant. This is because the mass does not affect the frequency of the oscillations, which is the number of cycles per unit time. However, a heavier mass will require a greater force to stretch or compress the spring, resulting in a larger amplitude of motion.

Can an inclined plane be used to measure an object's weight?

Yes, an inclined plane can be used to indirectly measure an object's weight by measuring its acceleration down the slope. Using the equation F = ma, where F is the force of gravity and m is the mass of the object, the weight can be calculated by multiplying the mass by the acceleration due to gravity (9.8 m/s^2). This method is based on the assumption that there is negligible friction on the plane.

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