Spring constant and velocity

In summary, a 3.2-kg block suspended from a spring with a force constant of 600 N/m and an unstretched length of 0.72 m is pulled down by an external force until the spring is stretched to 0.86 m. When the external force is removed, the block rises and when the spring contracts back to its original length, the upward velocity of the block is 0.96 m/s. This is found by using the conservation of energy equation, taking into account elastic and gravitational potential energy. The actual answer may be rounded to 1 decimal place.
  • #1
Yam
32
1

Homework Statement


[/B]
The force constant of a spring is 600 N/m and the un-stretched length is 0.72 m. A 3.2-kg block is suspended from the spring. An external force slowly pulls the block down, until the spring has been stretched to a length of 0.86 m. The external force is then removed, and the block rises. In this situation, when the spring has contracted to a length of 0.72m , the upward velocity of the block is = ?

Homework Equations


Conservation of Energy
Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

The Attempt at a Solution



0 + 0.5(600)(0.14)^2 = 0.5 x 3.2 x v^2 + 0
v = 1.92m/s?

This is incorrect. The four options given are
a) 1.0 m s b) 5.1m s c) 7.4 m s d) 9.1 m s e) 9.7 m sDo I need to account for gravitational potential energy in the equation?
[/B]
 
Last edited:
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  • #2
I think you may have mis-written your "Relevant Equation" , then when you substitute the numbers you got a bit mixed up.

Perhaps you could state each potential energy separately.

Edit: -oops, sorry I didn't read your final comment. Can you link to the original Q please?
 
  • #3
Hi, sorry for the mistake.
 
  • #4
Are you saying,
PE1 +KE1 = PE2 +KE2
0 + 0.5(600)(0.14)^2 = 0.5 x 3.2 x v^2 + 0

If so, a) when or where is "Initial" and when is "Final" ?
b) looking at your formula, one of your expressions is for KE1 and the other for PE2, but both look like 0.5 m v2
(though, what mass is 600?)

Perhaps you should state clearly where you are for each expression before you put it into an equation.

You might also think about what sorts of PE exist.
 
  • #5
PE1 = Elastic potential energy = 0.5 * k * x^2 =0.5(600)(0.14)^2
KE1 = 0

PE1 = gravitational potential energy = mass * g *height = (3.2)*(9.8)*(0.14)
KE2 = 0.5*m*v^2 = 0.5 x 3.2 x v^2

So,

0.5(600)(0.14)^2 = (3.2)*(9.8)*(0.14) + 0.5 x 3.2 x v^2

v = 0.96m/s?
 
  • #6
That looks ok now.
Presumably they've rounded the answer to 1dp.
 
  • Like
Likes Yam
  • #7
Thank you for your help! Stating clearly for each expression before putting into an equation really helps!
 

1. What is the relationship between spring constant and velocity?

The spring constant and velocity have an inverse relationship. This means that as the velocity of an object attached to a spring increases, the spring constant decreases, and vice versa.

2. How does the spring constant affect the velocity of an object?

The spring constant determines the rate at which the spring will stretch or compress in response to a force. This, in turn, affects the velocity of an object attached to the spring as it moves back and forth.

3. Can the spring constant change in different situations?

Yes, the spring constant can vary depending on the material and dimensions of the spring. It can also change if the spring is stretched beyond its elastic limit, causing it to become permanently deformed.

4. How is the spring constant measured?

The spring constant is typically measured in units of force per unit of length, such as Newtons per meter (N/m) or pounds per inch (lbs/in). It can be determined experimentally by measuring the force applied to the spring and the resulting displacement.

5. How does spring constant affect the force applied to an object?

The spring constant is directly proportional to the force applied to an object attached to the spring. This means that as the spring constant increases, the force applied to the object also increases, causing it to accelerate at a higher velocity.

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