**Problem Statement:**

A mass m = 5 kg starts its motion at point A (see figure) at a height h_{A} = 2 m with a speed v_{A} = 20 m/s. When it reaches point B (see figure) at a height h_{B} = 1.2m, its speed is v_{B} = 10 m / s.

- Determine the value of the work of the friction force between points A and B.
- If the radius of curvature of the track at point B is ρ = 2 m, check that the speed of the mass at this point is greater than the minimum required to take off.

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**Solution:**

To calculate the work of the friction force we will first determine the mechanical energy of the mass at points A and B. The difference between the two is the work of the friction force:

In the figure you can see that we are going to choose the zero height position at the lowest point of the trajectory.

Substituting the numerical values provided in the statement in the equation above with we get:

We have used g = 10 m/s^{2}

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In point (b) we will calculate the minimum speed necessary for the mass to take off at point B. We first draw the forces acting on it.

As you can see in the figure, the mass has a normal acceleration a_{n} at point B because the track is curved at this point and the velocity changes direction.

We have also represented a tangent axis and a perpendicular (normal) axis to the track at point B. These are the axes that we will use to project the vectors of Newton’s second law.

Newton’s second law for the mass at point B is:

And projecting onto the t and n axes we get:

Equation (1) would allow us to calculate the value of the tangential acceleration of the mass as a function of the frictional force. We are going to use equation (2).

The normal acceleration magnitude is given by:

Where ρ is the **radius of curvature** of the trajectory at point B.

After substituting this value in equation (2) we obtain:

When the particle takes off, it stops being in contact with the track. Therefore at this moment the **normal force is canceled**:

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

The problem statement tells us that the speed of the mass at point B is v_{B} = 10 m/s, this speed is therefore greater than the minimum value for the mass to take off.

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

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