Hooke's Law using Potential Energy

In summary, Hooke's Law describes the relationship between the force exerted by a spring and its displacement from the equilibrium position, stating that the force is proportional to the displacement. This principle can be connected to potential energy, as the work done on a spring when it is stretched or compressed is stored as elastic potential energy. The potential energy (U) in a spring is given by the formula U = (1/2)kx², where k is the spring constant and x is the displacement. This illustrates how mechanical energy is conserved in systems involving springs, as the energy can be converted between kinetic and potential forms.
  • #1
JMAMA
18
0
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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  • #2
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…
 
  • #3
haruspex said:
Not with friction present.
Max speed occurs when the acceleration…
when acceleration = 0
 
  • #4
JMAMA said:
when acceleration = 0
right, so what is the net force then? What are the horizontal forces on the box?
 
  • #5
F net= 0
F (friction) = F (spring)
 
  • #6
JMAMA said:
F net= 0
F (friction) = F (spring)
And what is the spring compression then?
 
  • #7
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
 
  • #8
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
 
  • #9
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.
 

FAQ: Hooke's Law using Potential Energy

What is Hooke's Law?

Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is expressed as F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.

How is potential energy related to Hooke's Law?

The potential energy stored in a spring due to its displacement is given by the formula U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement. This formula is derived by integrating the force over the distance.

What is the spring constant (k) and how is it determined?

The spring constant, denoted as k, is a measure of the stiffness of the spring. It is determined by the material properties and dimensions of the spring. The spring constant can be experimentally determined by measuring the force applied to the spring and the resulting displacement, then using Hooke's Law (k = F/x).

How do you calculate the work done in stretching or compressing a spring?

The work done on a spring when it is stretched or compressed is equal to the change in its potential energy. Thus, the work done W is given by W = (1/2)kx^2, where k is the spring constant and x is the displacement. This is the same as the potential energy stored in the spring.

Can Hooke's Law be applied to all types of materials and deformations?

No, Hooke's Law is only valid for materials and deformations where the relationship between force and displacement is linear. This typically applies to elastic materials within their elastic limit. Beyond this limit, materials may exhibit plastic deformation, and Hooke's Law no longer applies.

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