How Is Maximum Speed Calculated for a Block on a Vertical Spring?

In summary, the problem involves a spring with a spring constant of 40 N/m and an unstretched length of L0. A block of mass 1 kg is hung on the end of the spring and then pulled down until the spring is stretched twice the amount found in part (a). The block is then pushed upward with an initial speed of 4 m/s. The maximum speed of the block is to be determined using the equations for kinetic and potential energy. However, the incorrect value of 5.06 m/s is obtained due to not taking into account the gravitational potential energy. The correct equation to use is KEmax - KE0 = PE0 - PEmax, which takes into account the change in potential energy caused
  • #1
Dauden
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0

Homework Statement



https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/spring/homework/Ch-08-GPE-ME/mass_vertical_spring/7.gif

A spring with spring constant k = 40 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 1 kg is hung gently on the end of the spring.


Now the block is pulled down until the total amount the spring is stretched is twice the amount found in part (a). The block is then pushed upward with an initial speed vi = 4 m/s.

What is the maximum speed of the block?

Homework Equations


KE = (1/2)mv^2
PE = (1/2)kx^2

The Attempt at a Solution



Well, the answer for part a) they want me to use is .245 m. The way I set this up was:

(1/2)mv^2 = (1/2)mvi^2 + (1/2)kx^2

(1/2)(1)v^2 = (1/2)(1)(4^2) + (1/2)(40)(.49^2)

I did this because the kinetic energy at the maximum speed point should be equal to the potential plus the kinetic energy given by the push. When I solve for v on the left side I get 5.06 m/s but that is wrong.
 
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  • #2
Hi Dauden! :smile:
Dauden said:
Well, the answer for part a) they want me to use is .245 m. The way I set this up was:

(1/2)mv^2 = (1/2)mvi^2 + (1/2)kx^2

(1/2)(1)v^2 = (1/2)(1)(4^2) + (1/2)(40)(.49^2)

I did this because the kinetic energy at the maximum speed point should be equal to the potential plus the kinetic energy given by the push. When I solve for v on the left side I get 5.06 m/s but that is wrong.

No, KEmax - KE0 = PE0 -PEmax

and you haven't mentioned gravity :wink:
 
  • #3


I would approach this problem by first identifying the known variables and the relevant equations to use. In this case, the known variables are the mass of the block, the spring constant, the initial speed, and the maximum stretch of the spring. The relevant equations to use are the equations for kinetic energy and potential energy.

Next, I would carefully set up the equations and plug in the known values to solve for the maximum speed of the block. It is important to pay attention to units and use the correct values in the equations.

After getting a numerical answer, I would double check my calculations and make sure they make sense in the context of the problem. If the answer seems incorrect, I would go back and check my equations and calculations to find any mistakes.

Lastly, I would consider the significance of the answer and any potential sources of error. In this case, the maximum speed of the block is important because it tells us the maximum speed the block will reach after being pushed upwards. However, there may be some sources of error in the calculations, such as neglecting air resistance or assuming a perfectly ideal spring. These potential errors should be acknowledged and considered when interpreting the results.
 

Related to How Is Maximum Speed Calculated for a Block on a Vertical Spring?

1. What is a mass on a vertical spring?

A mass on a vertical spring refers to a system where a mass is attached to a spring that is fixed vertically to a surface. The spring allows the mass to oscillate up and down due to the force of gravity.

2. What is the relationship between the mass and the spring in this system?

The mass and the spring are connected through Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In other words, the heavier the mass, the greater the force exerted on the spring, and the more it will stretch or compress.

3. How does the amplitude of oscillation change when the mass is increased?

The amplitude of oscillation, which is the maximum displacement from the equilibrium position, increases with an increase in mass. This is because a heavier mass will exert a greater force on the spring, causing it to stretch or compress more, resulting in a larger amplitude of oscillation.

4. Does the length of the spring affect the oscillation of the mass?

Yes, the length of the spring does affect the oscillation of the mass. According to Hooke's Law, the force exerted by a spring is also directly proportional to the length of the spring. Therefore, a longer spring will have a lower force constant and will result in a slower oscillation compared to a shorter spring.

5. What factors affect the frequency of oscillation in this system?

The frequency of oscillation is affected by the mass, spring constant, and the length of the spring. A heavier mass, a stiffer spring, and a shorter spring length will result in a higher frequency of oscillation. Additionally, the amplitude of the oscillation also affects the frequency, with larger amplitudes resulting in slightly longer periods of oscillation.

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