Hooke's Law using Potential Energy

[JMAMA]

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

 

[haruspex]

Max speed occurs when all energy has been translated from spring into box.

Not with friction present.
Max speed occurs when the acceleration…

 

[JMAMA]

Not with friction present.
Max speed occurs when the acceleration…

when acceleration = 0

 

[haruspex]

when acceleration = 0

right, so what is the net force then? What are the horizontal forces on the box?

 

[JMAMA]

F net= 0
F (friction) = F (spring)

 

[haruspex]

F net= 0
F (friction) = F (spring)

And what is the spring compression then?

 

[JMAMA]

And what is the spring compression then?

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]

-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

 

[haruspex]

W (spring) = 1/2kx^2 = 0.14847855 Nm

Is that the energy the spring has released or the energy it has remaining?

 

[JMAMA]

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

 

[haruspex]

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.

 

[JMAMA]

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

 

[haruspex]

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.

 

[JMAMA]

You are forgetting work done against friction.

So PE (total) = Ke - W (friction) = 1/2k(x1^2 - x2^2) - (uk)mgx

 

[haruspex]

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.

 

[JMAMA]

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)

 

[JMAMA]

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

 

[haruspex]

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.

 

[JMAMA]

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

 

[haruspex]

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.

 

[JMAMA]

One of #16 and #17 is correct. Try to reason which one.

16 should be correct since 17 has friction doing positive work

 

[haruspex]

16 should be correct since 17 has friction doing positive work

Right.