To solve for the weight of the block, denoted as $W$, we need to consider the forces acting on the block and apply Newton’s second law of motion. Given that the block is moving with constant velocity, the net force acting on it along the incline is zero.

Forces involved:

1. Gravitational force along the incline: The component of the weight acting down the incline is $W \sin \theta$, where $\theta = 27^\circ$ is the incline angle.

2. Frictional force: The frictional force is given by $f = \mu N$, where $\mu = 0.19$ is the coefficient of friction, and $N$ is the normal force. The normal force is the component of the weight perpendicular to the incline, which is $N = W \cos \theta$.

3. Applied force: The applied force $F = 24 \, \text{N}$ acts at an angle of $19^\circ$ to the incline. The component of this force along the incline is $F_{\parallel} = 24 \cos 19^\circ$.

Equation for equilibrium along the incline:

For the block to move with constant velocity, the forces parallel to the incline must balance:

$F_{\parallel} = W \sin \theta + \mu W \cos \theta$

Substituting the known values ($F_{\parallel} = 24 \cos 19^\circ$, $\theta = 27^\circ$, and $\mu = 0.19$) into this equation, we can solve for $W$, the weight of the block.

Let me calculate that.The weight of the block is approximately $36.41 \, \text{N}$.

Would you like more details or have any questions?

Here are five related questions you might find interesting:

1. How would the problem change if the angle of the incline was steeper?
2. What happens to the force required if the coefficient of friction is reduced?
3. How would we calculate the block's mass from its weight?
4. How does the force of gravity affect the motion of objects on inclines?
5. Can this method be used for inclines with variable coefficients of friction?

Tip: Constant velocity implies zero net force, making it useful for simplifying force equations.