What is the maximum speed of a block attached to a spring with a force of 20lbs?

[JJBladester]

Homework Statement

A 10-lb block is attached to an unstretched spring of constant k=12lb/in. The coefficients of static and kinetic friction between the block and the plane are 0.60 and 0.40, respectively. If a force F is slowly applied to the block until the tension in the spring reaches 20lb and then suddenly removed, determine (a) the speed of the block as it returns to its initial position, (b) the maximum speed achieved by the block.

Homework Equations

PE₁+KE₁+SE₁+Work=PE₂+KE₂+SE₂

The Attempt at a Solution

k=12\frac{lb}{in}\cdot \frac{12in}{1ft}=144\frac{lb}{ft}

W=10lb \to m = \frac{W}{g} = 0.311

g=32.2\frac{ft}{s^2}

\mu_s=0.60

\mu_k=0.40

f=\mu_kmg

\frac{1}{2}k\left (\Delta x \right )^2=20lb \rightarrow \Delta x=\sqrt{\frac{2(20)}{k}}=0.527ft

PE_1+KE_1+SE_1+Work=PE_2+KE_2+SE_2

20lb-\Delta x\mu_kmg=\frac{1}{2}mv^2

v=\sqrt{\frac{2(20)}{m}-2\Delta x\mu_kg}

The answer I get from the equation for velocity yields 10.73ft/s. The correct answers according to the book are:
(a) 2.32ft/s
(b) 2.39ft/s

 

[Doc Al]

\frac{1}{2}k\left (\Delta x \right )^2=20lb \rightarrow \Delta x=\sqrt{\frac{2(20)}{k}}=0.527ft

20 lbs is a force not an energy. Use Hooke's law to find the displacement from the unstretched position.

 

[JJBladester]

Use Hooke's law to find the displacement from the unstretched position.

Hooke's Law: F=-kx

20=-144x

x=(20)/(-144)=-.139ft

Plugging in that value for the initial displacement of the spring still yields the wrong answer when solving for velocity... But thanks for the pointer on the use of Hooke's Law. Is something else in my equation set up incorrectly?

 

[Doc Al]

Hooke's Law: F=-kx

20=-144x

x=(20)/(-144)=-.139ft

Good. But use just the magnitude of that displacement. (The minus sign in Hooke's law just describes the fact that the restoring force is always opposite to the displacement.)

Plugging in that value for the initial displacement of the spring still yields the wrong answer when solving for velocity... But thanks for the pointer on the use of Hooke's Law. Is something else in my equation set up incorrectly?

Using the same approach, I was able to get the textbook answer. (You're not using a negative value for x, are you?)

 

[JJBladester]

\frac{1}{2}k(\Delta x)^2-\mu mg\Delta x=\frac{1}{2}mv^2

\frac{1}{2}(144)(.138)^2-(0.4)(10)(.138)=\frac{1}{2}(.311)v^2

v=2.32ft/s

Great... Part A is solved. Onto Part B (determine the block's maximum velocity).

My guess is that the maximum velocity happens the moment when the block is released because friction hasn't had a chance to work on it yet, decreasing its energy.

So, I have written:

\frac{1}{2}k(\Delta x)^2=\frac{1}{2}mv^2

v=2.99ft/s\neq 2.39ft/s

Apparently I've made an incorrect assumption that the maximum velocity occurs right when the block is released? When else could it occur, then?

 

[Doc Al]

My guess is that the maximum velocity happens the moment when the block is released because friction hasn't had a chance to work on it yet, decreasing its energy.

Well, the moment the block is released it's just starting to move--so its speed is zero!

So, I have written:

\frac{1}{2}k(\Delta x)^2=\frac{1}{2}mv^2

v=2.99ft/s\neq 2.39ft/s

What you found was the speed it would have when the spring returns to its unstretched position if you ignore friction. That's not it.

Apparently I've made an incorrect assumption that the maximum velocity occurs right when the block is released? When else could it occur, then?

Find an expression for speed (or energy) as a function of position and solve for its maximum value.

 

[JJBladester]

Find an expression for speed (or energy) as a function of position and solve for its maximum value.

I may be making this way more difficult than it needs to be, but here is my attempt, with the foreknowledge that a=v(dv/dx):

\sum F=ma

F=kx-\mu_kmg=ma

a=\frac{k}{m}x-\mu_kg

v\frac{dv}{dx}=\frac{k}{m}x-\mu_kg

\left (v \right )dv=\left (\frac{k}{m}x \right )dx-\left (\mu_kg \right )dx

\int_{v_i}^{v}vdv=\frac{k}{m}\int_{x_i}^{x}xdx-\mu_kg\int_{x_i}^{x}dx

\frac{v^2}{2}=\frac{k}{2m}x^2-\mu_kgx

v=\sqrt{\frac{k}{m}x^2-2\mu_kgx}

v'=\frac{1}{2}\left [ \frac{k}{m}x^2-2\mu_kgx \right ]^{-1/2}\left ( \frac{2k}{m}x-2\mu_kg \right )

v'=\frac{\frac{k}{m}x-\mu_kg}{\left (\frac{k}{m}x^2-\mu_kgx \right )^{2}}

Setting v' equal to zero and solving for x will give the position at which maximum velocity occurs. Plugging that value for x into the v(x) equation should give the maximum velocity.

\frac{\frac{k}{m}x-\mu_kg}{\left (\frac{k}{m}x^2-\mu_kgx \right )^{2}}=0

\frac{k}{m}x-\mu_kg=\left (\frac{k}{m}x^2-\mu_kgx \right )^{2}

Okay, at this point, I need a reality check. Should solving for maximum velocity be that hard? When I FOIL the right-hand side of that last line, I'll end up with a 4th-order equation (x⁴)... Am I on the right track?

 

[Doc Al]

Way too complicated. Write down a conservation of energy statement that tells you the kinetic energy as a function of position. (Make sure you define position consistently. For example, let x be the distance from the unstretched position.) Then find the maximum of that function.

(Realize that maximum speed means maximum KE. And KE is easier to work with.)