Spring Constants: Solving Sliding Block Problem

In summary, when the mass 12.0 kg slides down a frictionless 37.0 degree incline, it gains the energy of 212.3 Joules. The energy goes into the spring, which will compress by a length of .00607 meters.
  • #1
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A block of mass 12.0 kg slides from rest down a frictionless 37.0 degree incline and is stopped by a strong spring with k=3.50e4 N/m. The block slides 3.00 m from the point of release to the point where is comes to a rest against the spring. When the block comes to rest, how far has the spring been compressed?

I really wanted to use 1/2kx^2 = W to get this, but I know I didn't have all of the values. I thought of using KE = 1/2 mv^2, but I don't know how to get velocity without a time value. So I ended up using an equation for the incline, g sin theta = acceleration, got 5.90 m/s^2 for that, plugged mass and acceleration into f=ma to get a force of 70.8 N, plugged values into W=F S Cos theta to get work as 170 J, and finally plugged values into 1/2kx^2 = W to get x as .0984, but this is wrong. For every other equation I try to use, I'm missing 2 values and don't know how to get either of them. Where did I go wrong?
 
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  • #2
You're overcomplicating.
Use conservation of energy.
When an object of mass m goes down a vertical distance (height) h, then it gains the energy E = mgh (g is gravitational acceleration).
This energy goes into the spring, which will thus compress by a length x, where E = kx. OK?
 
  • #3
But it's not exactly a vertical distance since it's on an incline, is it?
 
  • #4
Come on. You know the distance, and you know the angle...
 
  • #5
So I took the sin of 37 times 3 to get the height. I got 1.805. Then I multiplied that times 12.0 and 9.8 to get E. I got 212.3. Then I divided that by 3.50e4. Got .00607. And it isn't right. I've been doing physics since 8 AM so I'm really tired, sorry. What am I missing?
 
  • #6
So I took the sin of 37 times 3 to get the height. I got 1.805.
Yes, that is the correct height (in meters).

Then I multiplied that times 12.0 and 9.8 to get E. I got 212.3.
Yes, that is the change in potential energy (in Joules).

Then I divided that by 3.50e4. Got .00607.
Why? Did you forget the formula 1/2kx^2 = W that you gave before?

x2= 2W/k or, using the figures you got
x2= 2(212.3)/(35000). Solve for x.
 
  • #7
thank you!
 

1. What is a spring constant?

A spring constant, also known as a force constant, is a measure of the stiffness of a spring. It is represented by the symbol k and is calculated by dividing the force applied to the spring by the displacement of the spring from its equilibrium position.

2. How is a spring constant used in the sliding block problem?

In the sliding block problem, the spring constant is used to calculate the force exerted by the spring on the block. This force is then used to determine the acceleration of the block and its motion.

3. What factors affect the spring constant?

The spring constant is affected by the material and shape of the spring, as well as the number of coils and the cross-sectional area of the spring. Additionally, the temperature and length of the spring can also have an impact on the spring constant.

4. How do you solve for the spring constant in the sliding block problem?

To solve for the spring constant, you will need to know the force applied to the spring, the displacement of the spring, and the mass of the block. Then, you can use the formula k = F/x to calculate the spring constant.

5. Can the spring constant change over time?

Yes, the spring constant can change over time due to factors such as wear and tear on the spring, changes in temperature, or changes in the material properties of the spring. It is important to regularly check and recalibrate the spring constant to ensure accurate results.

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