**Problem Statement:**

A block of mass m = 15 kg ascends with a constant velocity on an inclined plane with an angle α = 30º with respect to the horizontal. The coefficient of kinetic friction between the plane and the mass is μ_{k} = 0.2. Determine the magnitude of the force **F** that acts on the block.

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**Solution:**

To solve this problem we apply Newton’s second law. First we draw the forces acting on the block. As the force **F** causes the block to rise along the inclined plane, the friction force opposes the relative motion of this block with respect to the plane, so it has been drawn in the opposite direction to the velocity vector.

The normal (because the block is supported by the plane) and the weight (assuming that the block is close to the Earth’s surface) also act on the block.

We have also represented in the figure the reference frame that will be used to project Newton’s second law.

We have represented the projections of the weight vector on the Cartesian axes in the following figure:

The velocity of the block is constant, so the acceleration of the block is zero and Newton’s second law is:

The projections on the axes are:

From equation (2) we obtain the normal magnitude:

Note that **the normal and the weight do not have the same magnitude**.

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Since the block is moving, the friction force magnitude is given by:

Now we can substitute the value of the normal and the friction force in equation (1) and after substituting the givens we obtain:

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

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

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