Distance spring decompresses when friction is involved

In summary, a block of mass 4.30 kg traveling with a speed of 8.50 m/s hits a spring with a spring constant of 68.00 N/m and experiences a friction force with a coefficient of friction of 0.200. To find the distance the spring compressed when the block momentarily comes to rest, the work done by the system (friction and spring forces) is calculated using the equation Kf-Ki=W, where K represents kinetic energy and W represents work. The initial kinetic energy is found to be 155J, and the final kinetic energy is assumed to be 0. Solving for W, it is found to be -155J. Then, using the equation W=Fd (
  • #1
becky_marie11
8
0

Homework Statement


A block of mass m = 4.30 kg slides along a horizontal table with speed v0 = 8.50 m/s. At x = 0 it hits a spring with spring constant k = 68.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by μ = 0.200. How far did the spring compress when the block first momentarily comes to rest?
m=4.30 kg
vo=8.50m/s
k=68.00N/m
μ=.200
x=?

Homework Equations


Kf-Ki=W
K=1/2mv2
Fs=-kx
Ffric=μNf
W=Fd

The Attempt at a Solution


First I found the work of the whole system (friction and spring forces) by using
Kf-Ki=W.
For the initial kinetic energy, I used
K=1/2mv2=155J.
For the final kinetic energy, I used 0, because I'm measuring the distance after the mass came to a stop. So, plugging the numbers into the Work equation, I simply get
W=-155J
I'm pretty sure I'm on the right track so far?
Then, using
W=Fd (just multiplying the forcexdistance because both vectors are in the same direction)
I plugged in:
W=(-kx+μNf)x
I work it out to a quadratic by plugging the numbers in...
-68x2+8.44366x+155=0
I solve then for x (using the quadratic root finder on my calculator, so I'm not screwing up the quadratic..), but both of the answers it gives me are incorrect..
What am I doing wrong? Help please!
 
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  • #2
becky_marie11 said:
Then, using
W=Fd (just multiplying the forcexdistance because both vectors are in the same direction)
Careful: That only works when the force is constant. The spring force is not constant.

Hint: What's the energy stored in a compressed spring?

Also: Careful with signs. The work done on the mass by both the spring and the friction force is negative.
 
  • #3
welcome to pf!

hi becky! welcome to pf! :smile:

isn't it 1/2 kx2 ? :wink:
 
  • #4


tiny-tim said:
hi becky! welcome to pf! :smile:

isn't it 1/2 kx2 ? :wink:

W = ∫Fdx (F dot dx, technically)
Fspring = kx

W = ∫kxdx
Wspring = kx2/2

Just to show where it came from, since W does NOT equal Fx, that is not a definition, it is a derivation of work for a constant force.
 
Last edited:
  • #5
Thank you guys! I finally figured out the right answer! But when I solve for the quadratic equation, why is the distance vector only distributed to the friction force? This is what I tried to do...
W=(-1/2kx^2+friction force)*x
and this is what is right...
W=-1/2kx^2+friction force*x
Is it because the work of the spring is already in terms of "work" and the friction force isn't?
 
  • #6
hi becky! :smile:
becky_marie11 said:
This is what I tried to do...
W=(-1/2kx^2+friction force)*x
and this is what is right...
W=-1/2kx^2+friction force*x
Is it because the work of the spring is already in terms of "work" and the friction force isn't?

yes! :smile:

work done is force "dot" distance, and 1/2 kx2 is 1/2 kx (average force) times x (distance)! :wink:
 
  • #7
Thank you!
 

1. How does friction affect the distance a spring decompresses?

Friction can cause the spring to lose some of its potential energy as it decompresses, resulting in a shorter distance compared to a spring without friction. This is because the frictional force opposes the motion of the spring, causing it to slow down and compress less.

2. Is the distance a spring decompresses affected by the type of friction present?

Yes, the type of friction present can affect the distance a spring decompresses. For example, static friction, which occurs when two surfaces are in contact but not moving relative to each other, can have a greater impact on the distance compared to kinetic friction, which occurs when two surfaces are in motion relative to each other.

3. Can the distance a spring decompresses be calculated when friction is involved?

Yes, the distance a spring decompresses can be calculated when friction is involved. This calculation may be more complex compared to a spring without friction, as it requires taking into account the frictional force and its effect on the spring's motion. However, with the appropriate equations and data, the distance can be determined.

4. How can friction be minimized when studying the decompression of a spring?

To minimize the effects of friction on the decompression of a spring, it is important to use a smooth and clean surface for the spring to slide on. Additionally, reducing the weight and resistance of the spring can also help minimize the impact of friction. Lubricants or other techniques can also be used to reduce friction, but they must be carefully chosen to not alter the behavior of the spring.

5. Are there real-life applications where friction plays a significant role in the decompression of a spring?

Yes, there are many real-life applications where friction plays a significant role in the decompression of a spring. For example, in car suspension systems, friction can affect the compression and decompression of springs, which can impact the overall performance of the vehicle. In medical devices, such as prosthetics, the frictional forces between springs and joints can impact the movement and comfort of the device for the user.

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