Hooke's Law using Potential Energy

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Homework Help Overview

The discussion revolves around the application of Hooke's Law and potential energy in a system involving a spring and a box. Participants explore the relationship between potential energy stored in the spring and kinetic energy of the box, particularly in the context of frictional forces affecting the system's dynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the maximum speed of the box when energy is transferred from the spring, questioning the effects of friction and acceleration on this speed. There are attempts to calculate spring compression and work done by friction, with some participants suggesting algebraic approaches and others expressing confusion over the equations used.

Discussion Status

The conversation is ongoing, with participants providing various equations and reasoning about potential energy, kinetic energy, and work done against friction. Some guidance has been offered regarding the correct application of energy conservation principles, but there is no clear consensus on the correct equations or final outcomes.

Contextual Notes

Participants are navigating through assumptions related to friction and energy conservation, with some expressing uncertainty about the correct formulation of the equations involved. There is an emphasis on ensuring clarity in the definitions of potential and kinetic energy in the context of the problem.

JMAMA
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Homework Statement
A small box with mass 1.20 kg is placed against a light spring that is compressed 0.280 m . The spring, whose other end is attached to a wall, has force constant k = 42.0 N/m . The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. What is the maximum speed of the box during its motion?
Relevant Equations
E (kinetic) = 1/2mv^2
E (Potential) = 1/2kx^2
F (friction) = u m g
Max speed occurs when all energy has been translated from spring into box.
E (Potential) = 1/2kx^2
E (Potential) = (1/2)(42 N/m)(0.280 m)^2 = 1.6464 N m

Ep = Ek =1/2mv^2
1.6464 N m= 1/2 (1.2 kg) v^2
v = 1.6565 m/s
 
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JMAMA said:
Max speed occurs when all energy has been translated from spring into box.
Not with friction present.
Max speed occurs when the acceleration…
 
haruspex said:
Not with friction present.
Max speed occurs when the acceleration…
when acceleration = 0
 
JMAMA said:
when acceleration = 0
right, so what is the net force then? What are the horizontal forces on the box?
 
F net= 0
F (friction) = F (spring)
 
JMAMA said:
F net= 0
F (friction) = F (spring)
And what is the spring compression then?
 
haruspex said:
And what is the spring compression then?
haruspex said:
And what is the spring compression then?
-kx = uk m g
-X = [(0.3)(1.2 kg)(9.81 m/s2)]/42
X = -0.0840857143

Compressed by 0.084 m from equilibrium
 
JMAMA said:
-kx = uk m g
-X = [(0.3)(1.2 kg)(9.81 m/s2)]/42
X = -0.0840857143

Compressed by 0.084 m from equilibrium
W (spring) = 1/2kx^2 = 0.14847855 Nm
W (friction) = Fx = 0.296957109 NM

W tot = W (friction) - W (spring)
W tot = 0.148478559 NM

W tot = 1/2 mv^2
V = 0.4974578022

This seems wrong the velocity of the box at equilibrium is greater
 
JMAMA said:
W (spring) = 1/2kx^2 = 0.14847855 Nm
Is that the energy the spring has released or the energy it has remaining?
 
  • #10
haruspex said:
Is that the energy the spring has released or the energy it has remaining?
Remaining...
So X would actually be 0.195914286 from 0.280 - 0.084 m
Where

W (spring) = 1/2kx^2 = 0.806 Nm
W (friction) = Fx = 0.692 NM

W tot = W (friction) - W (spring)
W tot = 0.114 NM

W tot = 1/2 mv^2
V = 0.43
 
  • #11
JMAMA said:
So X would actually be 0.195914286 from 0.280 - 0.084 m
Where

W (spring) = 1/2kx^2 = 0.806 Nm
No, that doesn’t work either. A spring compressed by ##x_1## has PE ##\frac 12kx_1^2##. If you reduce the compression to ##x_2## it has PE ##\frac 12kx_2^2##. The reduction is ##\frac 12k(x_1^2-x_2^2)##, not ##\frac 12k(x_1-x_2)^2##.

Btw, it is a really good idea to work algebraically, using just symbols, not plugging in numbers until the end. It has many advantages.
 
  • #12
haruspex said:
No, that doesn’t work either. A spring compressed by ##x_1## has PE ##\frac 12kx_1^2##. If you reduce the compression to ##x_2## it has PE ##\frac 12kx_2^2##. The reduction is ##\frac 12k(x_1^2-x_2^2)##, not ##\frac 12k(x_1-x_2)^2##.

Btw, it is a really good idea to work algebraically, using just symbols, not plugging in numbers until the end. It has many advantages.
ok so once we have obtained the Potential energy using Pe = 1/2k(x1^2 - x2^2) = 1.49792 Nm do we just set that equal to 1/2mv^2 so, v^2 = {k(x1^2 - x2^2)}/m

sqrt[(42 N/m){(0.28m^2)-(-0.084m^2)}/1.2]= 1.58 m/s
 
  • #13
JMAMA said:
ok so once we have obtained the Potential energy using Pe = 1/2k(x1^2 - x2^2) = 1.49792 Nm do we just set that equal to 1/2mv^2 so, v^2 = {k(x1^2 - x2^2)}/m

sqrt[(42 N/m){(0.28m^2)-(-0.084m^2)}/1.2]= 1.58 m/s
You are forgetting work done against friction.
 
  • #14
haruspex said:
You are forgetting work done against friction.
So PE (total) = Ke - W (friction) = 1/2k(x1^2 - x2^2) - (uk)mgx
 
  • #15
JMAMA said:
So PE (total) = Ke - W (friction) = 1/2k(x1^2 - x2^2) - (uk)mgx
I don’t think either of those equations are what you meant. Please check.
 
  • #16
haruspex said:
I don’t think either of those equations are what you meant. Please check.
aren't those the equations we just solved for I don't know what else it would be
Ke = Pe - W (friction)
Ke = Pe - F (friction)(x)
1/2mv^2 = 1/2k(x1^2 - x2^2) - (uk)(m)(g)(x)
 
  • #17
JMAMA said:
aren't those the equations we just solved for I don't know what else it would be
Ke = Pe - W (friction)
Ke = Pe - F (friction)(x)
1/2mv^2 = 1/2k(x1^2 - x2^2) - (uk)(m)(g)(x)
or is this equation wrong should it be PE = KE - W(friction)
so KE = PE + W(friction)
1/2mv^2 = 1/2k(x1^2 - x2^2) + (uk)(m)(g)(x)

and solving for v, v = 2.036 m/s
 
  • #18
JMAMA said:
aren't those the equations we just solved for I don't know what else it would be
Ke = Pe - W (friction)
Ke = Pe - F (friction)(x)
1/2mv^2 = 1/2k(x1^2 - x2^2) - (uk)(m)(g)(x)
Yes, but that's not what you wrote in post #14.
 
  • #19
haruspex said:
Yes, but that's not what you wrote in post #14.
right I mixed up Pe and Ke so is the equation in post #16 correct? I think my equation in post #17 is the correct solution
 
  • #20
JMAMA said:
right I mixed up Pe and Ke so is the equation in post #16 correct? I think my equation in post #17 is the correct solution
One of #16 and #17 is correct. Try to reason which one.
 
  • #21
haruspex said:
One of #16 and #17 is correct. Try to reason which one.
16 should be correct since 17 has friction doing positive work
 
  • #22
JMAMA said:
16 should be correct since 17 has friction doing positive work
Right.
 

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