Solve Spring Equilibrium Problem: (a) & (b) w/ Friction

In summary, the block reaches the equilibrium position the first time with zero velocity. The frictional force is 2.0 N and it requires energy to remove this force so that the block will reach the equilibrium position.
  • #1
cgotu2
6
0
Please help with this problem!

A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.
(a) Calculate the speed of the block as it passes through the equilibrium position if no friction is present.

(b) Calculate the speed of the block as it passes through the equilibrium position if a constant frictional force of 2.0 N retards its motion.

(c) What would be the strength of the frictional force if the block reached the equilibrium position the first time with zero velocity?

I got part (a) by by saying that 1/2mv^2=1/2kA^2 and my answer of 11.898 cm/s was correct. but i can't get (b) or (c). in (b) i said that the work done by friction (2*0.33) was equal to 1/2mv^2 - 1/2kA^2, but my answer is incorrect.

i have no clue how to do (c). please help me!
 
Physics news on Phys.org
  • #2
for part b, the work done by friction is negative
 
  • #3
okay i did that, and i got 11.86 cm/s, but it is still wrong (for b)
 
  • #4
cgotu2 said:
Please help with this problem!

A 1.5 kg block is attached to a spring with a spring constant of 1950 N/m. The spring is then stretched a distance of 0.33 cm and the block is released from rest.

it's stretched 0.33 centimeters?

make sure you use the right units, i guess..
 
  • #5
From the work-kinetic energy theorem we have that the work done by all of the forces on a system is equal to the change in i'ts kinetic energy or
[tex]W=\Delta K[/tex]
We can split the work done into the two basis types for conservative an non conservative forces
[tex]W_{nc} \ + \ W_c=\Delta K[/tex]
and since the work done by the conservative force is equal to the negative of the change in the sytem's potential energy or
[tex]W_c=-\Delta U[/tex]
we can change the previous relation to
[tex]W_{nc} = \Delta K \ + \ \Delta U[/tex]
In this case the non conservative force is friction.

The last question is actually quite easy if one can answer this question: How much energy must be removed from the system, by the frictional force, in order to bring the mass to a halt at the equilibrium position?
 
Last edited:

What is the spring equilibrium problem?

The spring equilibrium problem involves finding the position of an object attached to a spring that is in a state of balance or equilibrium. This means that the forces acting on the object are equal and opposite, resulting in no net force on the object.

How do you solve the spring equilibrium problem?

To solve the spring equilibrium problem, you must first identify all the forces acting on the object, including the spring force, gravity, and any applied forces. Then, you can use Newton's laws of motion and the equation F=kx (where F is the force, k is the spring constant, and x is the displacement of the object from equilibrium) to calculate the displacement of the object from equilibrium.

What is the role of friction in the spring equilibrium problem?

Friction is a force that opposes the motion of an object. In the spring equilibrium problem, friction can affect the equilibrium position of the object by adding an additional force that must be accounted for in the calculations. Friction can also cause the object to vibrate or oscillate around the equilibrium position.

How does the spring constant affect the spring equilibrium problem?

The spring constant (k) is a measure of the stiffness of the spring and affects the amount of force required to stretch or compress the spring. A higher spring constant means a stiffer spring and a greater force required to achieve a certain displacement. In the spring equilibrium problem, the spring constant is used in the calculation of the spring force (F=kx).

What factors can affect the equilibrium position in the spring equilibrium problem?

The equilibrium position in the spring equilibrium problem can be affected by factors such as the spring constant, the mass of the object, the magnitude and direction of any applied forces, and the presence of friction. These factors can cause the object to shift its equilibrium position and result in a different displacement from equilibrium.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
265
  • Introductory Physics Homework Help
Replies
2
Views
993
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
31
Views
1K
  • Introductory Physics Homework Help
Replies
27
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
905
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
813
Back
Top