In the figure, a block of mass m=12kg is released from rest on a frictionless inclined of angle θ=300. At the bottom of inclined plane, there is spring as shown in the figure. To compress the spring by 2cm, 270 N force is required. Suppose the block momentarily stops when it compresses the spring by 5.5cm. What is the speed of the block just as it touches the spring?

To solve this problem, we can approach it using the principles of energy conservation. Here's the breakdown:

### Given:
- Mass of the block \( m = 12 \, \text{kg} \)
- Angle of inclination \( \theta = 30^\circ \)
- The force required to compress the spring by 2 cm is 270 N.
- The spring is compressed by \( x = 5.5 \, \text{cm} = 0.055 \, \text{m} \) when the block stops momentarily.
- We need to find the speed of the block just as it touches the spring.

### Step 1: Calculate the spring constant ( \( k \) )
The force required to compress the spring by 2 cm is given as 270 N. From Hooke's Law:
\[
F = k \cdot x
\]
where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the compression distance. Using the given force and compression:
\[
270 = k \cdot 0.02
\]
Solving for \( k \):
\[
k = \frac{270}{0.02} = 13,500 \, \text{N/m}
\]

### Step 2: Use energy conservation
At the top of the inclined plane, the block has potential energy due to its height, and at the bottom (just as it touches the spring), all this potential energy is converted into kinetic energy.

At the moment when the block has compressed the spring by 5.5 cm, all the kinetic energy has been converted into spring potential energy, and the block comes to rest momentarily. Thus, we can set up the following equations:

- Gravitational potential energy at the top: 
  \[
  U_g = mgh
  \]
  where \( h \) is the vertical height.

- The vertical height \( h \) can be related to the distance traveled along the incline by the block, \( d \), as follows:
  \[
  h = d \sin \theta
  \]

- Kinetic energy just before the block touches the spring:
  \[
  K = \frac{1}{2} m v^2
  \]

- Spring potential energy when the block compresses the spring by 5.5 cm:
  \[
  U_s = \frac{1}{2} k x^2
  \]

### Step 3: Energy conservation equation
Initially, the block has potential energy, and when it reaches the spring, this energy is converted into kinetic energy. As the block compresses the spring, its kinetic energy is converted into spring potential energy. Using energy conservation:

\[
mgh = \frac{1}{2} m v^2 + \frac{1}{2} k x^2
\]

At the point just before touching the spring, the block has not yet compressed the spring, so the energy balance is between gravitational potential energy and kinetic energy:
\[
mgh = \frac{1}{2} m v^2
\]

Let's calculate \( h \). The height of the block from the inclined plane is given by the vertical component:
\[
h = d \sin \theta
\]
We can find \( d \) if we know how far the block moved along the inclined plane.

Now, if you can provide more details about the distance the block traveled, I can help you further with the calculations. Would you like to clarify this, or ask any questions so far? 

Here are some related questions for deeper understanding:
1. How does the compression of the spring relate to the block's potential energy?
2. What factors influence the spring constant in a real-world setup?
3. How does the angle of inclination affect the block's speed?
4. What is the significance of the energy conservation principle in this context?
5. How would friction on the inclined plane alter the outcome?

**Tip:** Always ensure all units are converted to the same system (e.g., meters, kilograms) to avoid calculation errors.

Learn how to calculate the speed of a 12kg block released from rest on a frictionless inclined plane with a 30° angle. Using energy conservation principles and Hooke's Law, this problem explores potential energy, kinetic energy, and spring compression when the block compresses the spring by 5.5cm.

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