Box on a string, find displacement when first at rest

In summary: When the block first comes to rest, it will move in the opposite direction because the spring exerts a sufficient force.
  • #1
lindzzz
4
0

Homework Statement


A block with mass m = 3.57 kg is attached to a spring with spring constant k = 165.7 N/m and negligible mass. The coefficient of kinetic friction between the block and the table is µk = 0.239. The box is displaced from its equilibrium position to the left by 0.216 m. If the block is released from rest, calculate the block's displacement when it first comes to rest again.
(Picture attached)

Homework Equations


Us = 1/2kΔL²
Wfriction = μmgdcosθ


The Attempt at a Solution



This is what I tried:
Us = Wfriction
1/2(k)(xf)² - 1/2(k)(xi)² = μmgdcosθ
I then said xf = d - xi and subbed in this for xf
1/2(165.7)(d-0.216)² - 1/2(165.7)(0.216)² = 0.239(3.57)(9.8)d(cos180°)
solving for d I get d = 0.33 m but this is incorrect
I'm not sure what I'm doing wrong, suggestions?
 

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  • #2
When the block first comes to rest, does the spring exert sufficient force to move the block in the opposite direction. If so, then friction hasn't yet dissipated all the energy initially stored in the spring, and your energy approach won't work. Try solving the equations of motion and calculating the time at which the block comes to rest.
 
  • #3
The reason I'm doing it with energy is that this problem is under the heading: Thermal energy and power. How I see it is if the ground were frictionless then the energy stored in the spring at xi would be equal to that stored at xf, and energy is conserved. However with friction energy will be lost to heat therefore the difference in energy stored in the spring at xf and xi should be equal to the energy lost to heat (friction).
 
  • #4
"The reason I'm doing it with energy is that this problem is under the heading: Thermal energy and power."

Got it. You'll have to account for the energy stored in the spring when the block comes to rest:

(Initial Spring Energy) = (Energy Dissipated) + (Spring Energy When Block Stops)

= μmgd + k(d - xi)^2

Actually, that's easier than solving the equations of motion!
 
  • #5


I would approach this problem by first understanding the physical principles involved. The box is attached to a spring, which means that it is experiencing a restoring force that is proportional to its displacement from its equilibrium position. This can be described by Hooke's Law, F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement from equilibrium.

In this case, the box is initially displaced from its equilibrium position by 0.216 m. As it is released from rest, it will experience a restoring force that will accelerate it towards its equilibrium position. However, the presence of friction will also act on the box, opposing its motion. This means that the box will eventually come to a stop at a certain displacement from its starting point.

To determine this final displacement, we can use the principle of conservation of energy. At the initial displacement of 0.216 m, the box has a certain amount of potential energy stored in the spring, given by Us = 1/2kx². As it moves towards its equilibrium position, this potential energy will be converted into kinetic energy, given by KE = 1/2mv². However, some of this kinetic energy will be lost due to friction, as given by Wfriction = μmgdcosθ.

At the point where the box comes to rest, all of its kinetic energy will have been converted into frictional work. This means that we can set the equations for potential energy and frictional work equal to each other, and solve for the final displacement, x.

1/2kx² = μmgdcosθ
x = √[(2μmgdcosθ)/k]

Plugging in the values given in the problem, we get x = √[(2*0.239*3.57*9.8*0.216*cos180°)/165.7] = 0.2 m. This is the final displacement of the box when it comes to rest again.

In summary, as a scientist, I would approach this problem by understanding the physical principles involved and using the appropriate equations to solve for the unknown variable. I would also carefully check my calculations and units to ensure accuracy in my answer.
 

1. What is a "box on a string" experiment?

The "box on a string" experiment is a simple physics experiment that involves a small box attached to a string. The string is pulled with a constant force and the box is allowed to move in a straight line. This experiment is used to study the concept of displacement and its relation to force and motion.

2. How do you find displacement in a "box on a string" experiment?

To find the displacement in a "box on a string" experiment, you need to measure the initial and final position of the box on the string. The displacement is then calculated by subtracting the initial position from the final position. This gives the total distance the box has moved in a straight line.

3. What factors affect the displacement in a "box on a string" experiment?

The displacement in a "box on a string" experiment is affected by the force applied to the string, the mass of the box, and the length of the string. The greater the force applied, the greater the displacement will be. Similarly, a heavier box or a longer string will result in a larger displacement.

4. Can the displacement be negative in a "box on a string" experiment?

Yes, the displacement can be negative in a "box on a string" experiment. This occurs when the final position of the box is behind the initial position. It is important to keep track of the direction of motion in this experiment to accurately calculate the displacement.

5. How is displacement different from distance in a "box on a string" experiment?

Displacement and distance are two different concepts in physics. Displacement refers to the straight-line distance between the initial and final positions of the box. Distance, on the other hand, refers to the total length of the path traveled by the box. In a "box on a string" experiment, displacement will always be equal to or less than the distance traveled.

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