Work Problem -- Wood Block Sliding into a Spring

AI Thread Summary
The discussion revolves around calculating work done on a wooden block sliding into a spring, focusing on the integration of forces involved. The work done by the spring and friction is calculated, resulting in -89.9 Joules, indicating that work is done on the block. There is confusion regarding the treatment of the spring constant and the direction of forces, with clarification that negative work occurs when the force and displacement are opposite. The integration of friction is typically treated as constant, and the net work can be derived without complex integration. The conversation emphasizes understanding the signs and directions of forces in work calculations.
Che8833
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Homework Statement
Suppose the force exerted by a spring is F(x) = -kx and the spring has constant k = 325 N/m. Then suppose a wooden cube headed east with mass m = 6 kg traveling with an initial velocity on a level surface with kinetic coefficient of friction 0.250 presses against the spring and it compresses from x = 0 to x = 0.70 meters as it stops. What is the net work done? HINT: You need to integrate to solve.
Relevant Equations
F(x) = -kx and the spring has constant k = 325 N/m

Kinetic coefficient of friction 0.250

Spring compresses from x = 0 to x = 0.70 meters
W_net = Integral from 0 to 0.70 meters [ - F_spring - F_friction ]

= 1/2 * (-k) * x^2 - mu * mg * normal force * x

= 1/2 * (-325N/m) * (.70)^2 - 0.250 * 6kg * 9.81m/s^2 * 0.70 - 0

= - 89.93 Joules

Is this correct and am I setting this whole thing up correctly? The negative signs have me bit confused. Also, am I supposed to integrate the Force of friction like this or do I treat it as a constant?

Thanks for the help
 
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You didn't show how you integrated the friction force, but yes, one usually assumes it's constant. You know what happens when you integrate a constant, right?

-89.9 J of work is done on the wooden cube.
+89.9 J of work is done by the wooden cube.

It's not clear which of these is being asked for, but I would guess the former.
 
Mister T said:
You didn't show how you integrated the friction force, but yes, one usually assumes it's constant. You know what happens when you integrate a constant, right?

-89.9 J of work is done on the wooden cube.
+89.9 J of work is done by the wooden cube.

It's not clear which of these is being asked for, but I would guess the former.

Does it make sense that I am leaving the spring constant negative and subtracting the work done by friction? I was thinking that should be the case since its vector is going the same direction as the spring is pushing or am I wrong about this? I saw a similar problem online and the spring constant k was positive and the work done by friction was being subtracted. Can you please explain what's correct? Thanks for your help.
 
Che8833 said:
Does it make sense that I am leaving the spring constant negative and subtracting the work done by friction?

But you didn't make ##k## negative. You simply have a minus sign in front of ##k##. It makes no difference where you put it. ##-ab=a(-b)=(-a)b##.
 
Yes if you are talking the net work done on the block, both friction and the spring force are opposite the motion (displacement) of the block. Can you think of an example where the work done on the block by the spring is positive?
 
PhanthomJay said:
Yes if you are talking the net work done on the block, both friction and the spring force are opposite the motion (displacement) of the block. Can you think of an example where the work done on the block by the spring is positive?

If the block was moving west and the spring was pushing the block east to stop it, would the equation then be F(x) = kx or it would it still F(x) = -kx since the spring is slowing the block down? Because wouldn't it still be considered negative work?
 
Negative work occurs when the force and displacement of the object along its line of action are in opposite directions. So can you answer your last question?
 
Last edited:
Che8833 said:
What is the net work done? HINT: You need to integrate to solve.
Net work done on the block or net work done on the spring?
Either way, I cannot see any need to integrate. Each of those answers can be written down immediately using very standard equations.
 
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