**Problem Statement:**

A block of mass m = 15 kg is attached to a spring of stiffness K = 100 N/m. The block descends a plane inclined at an angle α = 30º with the horizontal. Assuming there is no friction, determine the acceleration of the block when the spring has stretched by a length x = 0.02m.

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**Solution:**

To solve this problem we are going to use Newton’s second law. The forces that act on the block are the weight (if it is near the Earth), the normal (because the block is supported by the plane) and the force of the spring that is given by Hooke’s law.

Next we will draw a diagram of the forces that act on the block along with the Cartesian axes that we will use to make the projections.

The return force of the spring acts in the opposite direction to the elongation of the spring (and consequently to the direction of motion of the mass):

The acceleration of the block has also been represented in the figure. It is directed in the positive direction of the *x-axis*.

The following diagram shows the projections of the weight vector on the chosen axes:

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Newton’s second law applied to the motion of the block is:

With the projection on the Cartesian axes we get:

From equation (2) we obtain the normal magnitude. As you can see, **is not equal to the weight**.

In addition, the magnitude of the spring recovery force is given by:

Note that although the vector force given by Hooke’s law has a minus sign, as we use here **its magnitude it must be positive**.

Finally, we can deduce the acceleration from equation (1) and after substituting the givens we get:

In the problem we have used g = 10 m/s^{2}

**Do not forget to include the units in the results of the problem.**

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