Hooke's Law using Potential Energy

  • #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.
 

1. What is Hooke's Law?

Hooke's Law is a fundamental principle in physics that describes the relationship between the force applied to an elastic material and the resulting deformation of that material. It states that the force applied is directly proportional to the displacement of the material.

2. What is potential energy in the context of Hooke's Law?

Potential energy in the context of Hooke's Law refers to the energy stored in an elastic material when it is deformed. It is a form of elastic potential energy and is directly related to the amount of displacement of the material.

3. How is Hooke's Law used to calculate potential energy?

To calculate potential energy using Hooke's Law, the formula PE = 1/2 * k * x^2 is used, where PE is the potential energy, k is the spring constant of the material, and x is the displacement of the material. This formula assumes that the force applied is constant and that the material is being stretched or compressed in a linear manner.

4. What is the significance of Hooke's Law in real-world applications?

Hooke's Law has many real-world applications, such as in the design of springs, shock absorbers, and other elastic materials. It is also used in engineering and construction to determine the strength and stability of structures under various forces. Additionally, it is used in medical devices, such as prosthetics and braces, to provide support and flexibility.

5. Are there any limitations to Hooke's Law?

While Hooke's Law is a useful concept, it does have some limitations. It only applies to materials that behave in a linear manner, meaning that the force applied is directly proportional to the displacement. It also assumes that the material will return to its original shape and size once the force is removed, which may not always be the case. In some cases, materials may exhibit hysteresis, where the amount of force needed to deform the material is different from the force needed to return it to its original shape. Additionally, Hooke's Law does not take into account factors such as temperature, which can affect the behavior of materials.

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