To find the x-component of the force acting on the block, we can use the formula:

$F_x = F \cdot \cos(\theta)$

Where:

• $F_x$ is the x-component of the force,
• $F = 177 \, \text{N}$ is the total force applied,
• $\theta = 85^\circ$ is the angle between the force and the horizontal.

Now, I'll calculate the x-component.The x-component of the force acting on the block is approximately $15.43 \, \text{N}$.

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the y-component of the force acting on the block?
2. How would the x-component change if the angle were 60° instead of 85°?
3. What is the resultant force on the block when considering both x and y components?
4. How do friction forces impact the movement of the block in this situation?
5. What if the block was pulled on an inclined plane—how would the components change?

Tip: Always use the cosine of the angle to calculate the horizontal (x) component and the sine for the vertical (y) component in force problems.