The problem asks to find the magnitude of the force $F$ required to push a block of mass $m = 15.0 \, \text{kg}$ up an incline at a constant velocity, where the incline has an angle $\theta = 30^\circ$, and the coefficient of kinetic friction is $\mu_k = 0.375$.

Breakdown of Forces:

1. Gravitational force $F_g = mg$, where $g \approx 9.8 \, \text{m/s}^2$.

   • The component of gravity parallel to the incline is $F_g \sin\theta = mg\sin\theta$.
   • The component of gravity perpendicular to the incline is $F_g \cos\theta = mg\cos\theta$.

2. Normal force $N$: Since there is no acceleration perpendicular to the incline, the normal force is balanced by the perpendicular component of gravity: $N = mg \cos\theta$

3. Frictional force $F_f$: The kinetic friction opposes the motion and is given by $F_f = \mu_k N$. Substituting $N$: $F_f = \mu_k mg \cos\theta$

4. Total force required $F$: To maintain constant velocity, the applied force $F$ must balance both the force due to gravity (parallel to the incline) and the frictional force. The equation for the required force is: $F = F_g \sin\theta + F_f$ Substituting the expressions for $F_g$ and $F_f$: $F = mg \sin\theta + \mu_k mg \cos\theta$

Now, let’s calculate the magnitude of $F$.The magnitude of the force $F$ required to push the block up the incline at a constant velocity is approximately $121.24 \, \text{N}$.

Would you like further details on any part of the solution, or have any specific questions?

Here are five related questions to expand your understanding:

1. How would the force change if the incline angle increased?
2. What happens to the frictional force if the surface becomes smoother (lower $\mu_k$)?
3. How would the required force change if the mass of the block doubled?
4. What would happen to the force if the block was moving down the incline at a constant velocity?
5. How does the constant velocity assumption simplify the problem?

Tip: In problems involving inclines, always break forces into components parallel and perpendicular to the incline.