The horizontal force \( F_P = 150 \, \text{N} \) pushes the block of 5.0 m up along the 32° incline. A coefficient of friction \( \mu_k = 0.13 \). [Hint: Work-energy involves net work done.]**Diagram Explanation:**- The diagram illustrates a block weighing 18 kg on a 32° inclined plane.- The force \( F_P = 150 \, \text{N} \) is applied horizontally in the direction of the slope.- The slope is represented with the angle 32° marked beside it.- The block is moving upward along the incline. **Part C****How much work is done by the normal force?***Express your answer to three significant figures and include the appropriate units.*\( W_N = \)- **Value**: [Enter the calculated numerical value here]- **Units**: [Enter the corresponding units here]This section involves calculating the work done by the normal force. To proceed, remember that the work done by a force is calculated as the dot product of the force vector and the displacement vector. In cases where the displacement is perpendicular to the force, such as with a normal force, the work done is typically zero.The interface includes:- An options panel with tools like formatting, symbols, and undo/redo buttons.- A keypad for entering numerical values and units.

Name: The horizontal force \( F_P = 150 \, \text{N} \) pushes the block of
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: The horizontal force \( F_P = 150 \, \text{N} \) pushes the block of 5.0 m up along the 32° incline. A coefficient of friction \( \mu_k = 0.13 \). [Hint: Work-energy involves net work done.]**Diagram Explanation:**- The diagram illustrates a block w