Sliding blocks on an inclined plane

[Fanta]

Homework Statement

Two blocks of mass m1 = 1kg and m2 = 2kg, connected through a rope, slide across an inclined plane with inclination of

$$\theta = \frac{\pi}{6}$$

The friction coefficient equals 0.5

3.1 Calculate the total work done by the friction force that acts on the first block after it has moved 1m along the inclined plane.

3.2 Determine the accelaration of the system and the rope tension.

Homework Equations

$$\theta = \frac{\pi}{6}$$

$$\mu = 0.5$$

$$F_{f} = \m \cdot u \cdot N$$

$$F = m \cdot a$$

g = 9.81

The Attempt at a Solution

3.1 -

To calculate the friction force in the first block, i think we don't have to consider the second block, so:

$$F_{f} = \mu \cdot N$$

$$N = m_{1} \cdot g \cdot cos \theta$$

$$F_{f} = \mu \cdot m_{1} \cdot g \cdot cos \theta$$

then we multiply that for the displacement (1m), and we get the work.

Is this correct?

3.2 Determine the acceleration of the system and the rope tension.

First, to determine the acceleration, I considered the whole system as a whole. So, consideing only external forces, I sum the masses 1 and 2, giving me M -> mass of the whole system.

$$Fr = F_{gx} - F{f}$$

$$M \times a = (M g sin \theta) - (M g cos \theta)$$

$$a = \frac{(M g sin \theta) - (M g cos \theta)}{M}$$

now, for the tension:

$$F{r1} = (m{1} g sin \theta) - (m{1} g cos \theta)$$

$$F{r2} = (m{2} g sin \theta) - (m{2} g cos \theta)$$

$$Tension = F{r2} - F{r1}$$

Is everything right?

 

[anathema]

I would like to clarify a few things in your solution.

3.1 - Your approach to calculating the work done by friction is correct. However, I would like to mention that the friction force will depend on the normal force, which in turn depends on the weight of both blocks. So, it is important to consider the weight of both blocks in your calculation.

3.2 - Your approach to determining the acceleration of the system is correct, but I would like to mention that the mass of the system should be the sum of both blocks, not just the external forces. So, the equation should be M = m1 + m2. Additionally, you have used the same value for g in both the numerator and denominator, which is incorrect. The correct equation should be:

a = (g sin theta - u g cos theta)/(m1 + m2)

For the tension, your approach is correct, but I would like to mention that the tension in the rope will be the same for both blocks, as they are connected by the same rope. So, the equation should be:

Tension = F{r2} = F{r1} = m1 g sin theta - m1 g cos theta

Overall, your approach is correct, but it is important to be careful with the values and equations used. I hope this helps in your solution.

 

[kerimek]

Overall, your approach and equations seem to be correct. However, there are a few points that can be clarified or improved upon.

3.1 - Your approach to calculating the work done by the friction force on the first block is correct. However, you should also consider the work done by the gravitational force on the first block, which is equal to m1gcosθ multiplied by the displacement (1m). The total work done on the first block would be the sum of these two values.

3.2 - Your approach to calculating the acceleration of the system is correct. However, you should also consider the tension in the rope when calculating the acceleration. The tension in the rope is equal to the sum of the forces acting on the second block (m2gcosθ) and the friction force on the first block (μm1gcosθ). This tension will also contribute to the acceleration of the system.

Additionally, for the tension calculation, you should use the net force on the second block (m2gcosθ) instead of the individual forces (m2gsinθ and m2gcosθ).

Overall, your approach is correct, but it would be helpful to include all relevant forces and their contributions to the system's acceleration and work done.