Spring Speed Problem: Solve 4kg Block, 3500N/m Stiffness

  • Thread starter Thread starter timmy8
  • Start date Start date
  • Tags Tags
    Speed Spring
Click For Summary

Discussion Overview

The discussion revolves around a physics homework problem involving a spring launching a block. Participants explore the application of energy conservation principles to determine the speed of a block at a specific height after being released from a compressed spring. The focus includes the calculation of potential energies and kinetic energy in the context of the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • The initial setup involves a spring with a stiffness of 3500 N/m and a 4 kg block, compressed by 0.2 m, with the goal of finding the block's speed at 1.3 m above its starting position.
  • Some participants propose using the conservation of energy equation: delta_Uspring + delta_Ugravitation + delta_K = 0.
  • One participant expresses confusion about their calculations, noting that they obtained a negative value for Vf^2, indicating a potential error in their approach.
  • Another participant suggests that the change in spring potential energy may have been calculated incorrectly and questions the signs used in the gravitational potential energy calculation.
  • Clarifications are made regarding the formulas for gravitational potential energy, with emphasis on the positive change in height as the block rises.
  • One participant confirms that they arrived at the correct answer after adjusting their equations, thanking another participant for their assistance.

Areas of Agreement / Disagreement

Participants generally agree on the need to apply conservation of energy principles, but there is some disagreement and confusion regarding the correct application of the formulas and the signs used in calculations. The discussion remains unresolved in terms of a definitive method for solving the problem, as participants are still refining their understanding.

Contextual Notes

Participants express uncertainty about the correct application of energy conservation principles, particularly regarding the signs in their calculations and the treatment of potential energies. There is a lack of consensus on the specific steps leading to the solution.

timmy8
Messages
5
Reaction score
0

Homework Statement


A spring whose stiffness is 3500 N/m is used to launch a 4 kg block straight up in the classroom. The spring is initially
compressed 0.2 m, and the block is initially at rest when it is released. When the block is 1.3 m above its starting
position, what is its speed?


Homework Equations


System: block, spring and earth.
delta_Uspring + delta_Ugravitation + delta_K=0


The Attempt at a Solution


everything I've tried seems to be wrong, the answer answer given to us however is 3.09 m/s.
 
Physics news on Phys.org
timmy8, welcome to PF!

Your equation is correct, so please show your work so we may see where you may have gone wrong.
 
1/2(3500)(1.3^2)-1/2(3500)(0.2^2)+4(-9.8)(1.3-0.2)+1/2(4)Vf^2-1/2(4)(0^2)=0

from there i solved for vf^2;

and vf^2=-1422.19

i am getting Vf^2 to be a negative number and i would have to solve the square root of that to get Vf.
what am i doing wrong?
 
timmy8 said:
1/2(3500)(1.3^2)-1/2(3500)(0.2^2)+4(-9.8)(1.3-0.2)+1/2(4)Vf^2-1/2(4)(0^2)=0

from there i solved for vf^2;

and vf^2=-1422.19

i am getting Vf^2 to be a negative number and i would have to solve the square root of that to get Vf.
what am i doing wrong?
You have the change in the spring potential energy wrong. What is the PE of the spring when the block rises 1.3 m above its start point? The block is not attached to the spring. Also, you seem to have slipped up on a plus /minus sign for the change in gravitational PE. Also, the problem asks for the speed at 1.3 m above the initial start position of the block (when the spring is initially compressed, I think).
 
so in the spring potential should Uf be equal to zero? and for the change in gravitational PE isn't the formula m*g*delta_h? I am so lost.
 
timmy8 said:
so in the spring potential should Uf be equal to zero?
yes
and for the change in gravitational PE isn't the formula m*g*delta_h? I am so lost.
Yes, the formula is change in Gravitational PE is (mgh_final) - (mgh_initial). final h is 1.3 and initial h is zero, so the change is positive (it has gained grav. PE).
 
i thought g would be -9.81?
 
now i have:
-springPE +mgh_final+1/2(m)v_final^2=0

This gave me the correct answer.

thank you for all your help PhantomJay
 
timmy8 said:
now i have:
-springPE +mgh_final+1/2(m)v_final^2=0

This gave me the correct answer.

thank you for all your help PhantomJay
You're welcome. The plus and minus sign can be a killer in Physics, and it is easy to get confused. Sometimes you just have to reason it out.
 

Similar threads

Replies
29
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 11 ·
Replies
11
Views
5K
Replies
8
Views
6K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K