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Spring compressed, find velocity.

  • #1

Homework Statement


A block of mass 0.3 kg and spring constant 24 N/m is on a frictionless surface. If the block is set into motion when compressed 3.5 cm, what is the maximum velocity of the block? How much is the spring compressed when the block has a velocity of 0.19 m/s?


Homework Equations


F = -kx


The Attempt at a Solution


m = 0.3 kg
k = 24 N/m
3.5cm = 0.035m is what? amplitude or distance I'm not sure.

F = 24*0.035 = 0.84 N?

I'm stuck, any help would be greatly appreciated :)
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi demonslayer42! :smile:

use conservation of energy (∆PE + ∆KE = constant) :wink:
 
  • #3
Delphi51
Homework Helper
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In F = -kx, x is the compression or stretch of the spring, so at first the force on the mass is F = k*0.035 = 0.84 N as you found. You could get the initial acceleration with F = ma. However, as the mass moves and the spring is less compressed, the force and acceleration decrease. Calculus integration would be necessary to deal with the diminishing acceleration.

This problem is easier to deal with using energy formulas.
The energy stored in a spring is Es = ½kx². You can answer the first question by starting with
Initial Es = kinetic energy at maximum speed
 
  • #4
I'm sorry, but I don't understand what you are saying.
Es = ½kx²
(1/2)(24)(0.035)^2 = 0.0147 ? So if this is Es what am I suppose to do with this?
 
  • #5
(1/2)mv^2 + (1/2)kx^2 = k

(1/2)(0.3)(v^2) + 0.0147 = 24

(1/2)(0.3)(v^2) = 23.9853

v^2 = 23.9853/0.15
=12.645 ?
That doesn't look right.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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hi demonslayer42! :smile:

(try using the X2 icon just above the Reply box :wink:)
(1/2)mv^2 + (1/2)kx^2 = k
no, wrong constant

your equation is KE + PE = constant,

but that shouldn't be the spring constant, it should be the constant that fulfils the initial conditions …

so set the constant at the value of KE + PE at time zero :smile:
 

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