---
title: "A block of mass \\(M\\) is pulled a distance \\(d\\) along a horizontal surface by a constant force of magnitude \\(F\\). In Trial 1, the force is applied horizontally in the direction of the displacement. In Trial 2, the force is applied at an angle \\(\\theta\\) above the horizontal, where \\(0^\\circ < \\theta < 90^\\circ\\). Which of the following correctly compares the work done by the applied force in Trial 1, \\(W_1\\), to the work done by the applied force in Trial 2, \\(W_2\\), and provides a valid justification?"
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url: "https://nerd-notes.com/ubq/113064/"
date_modified: "2026-05-05T03:45:15+00:00"
---

# A block of mass \(M\) is pulled a distance \(d\) along a horizontal surface by a constant force of magnitude \(F\). In Trial 1, the force is applied horizontally in the direction of the displacement. In Trial 2, the force is applied at an angle \(\theta\) above the horizontal, where \(0^\circ < \theta < 90^\circ\). Which of the following correctly compares the work done by the applied force in Trial 1, \(W_1\), to the work done by the applied force in Trial 2, \(W_2\), and provides a valid justification?

A block of mass \(M\) is pulled a distance \(d\) along a horizontal surface by a constant force of magnitude \(F\). In Trial 1, the force is applied horizontally in the direction of the displacement. In Trial 2, the force is applied at an angle \(\theta\) above the horizontal, where \(0^\circ < \theta < 90^\circ\). Which of the following correctly compares the work done by the applied force in Trial 1, \(W_1\), to the work done by the applied force in Trial 2, \(W_2\), and provides a valid justification?

![Two diagrams showing a block on a horizontal surface being pulled to the right. In Trial 1, a force vector F points horizontally to the right. In Trial 2, a force vector F points diagonally up and to the right at an angle theta relative to the horizontal surface.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1777952715-qoIAqR.jpg)

- **A.** \(W_1 > W_2\) because only the component of the applied force parallel to the displacement performs work on the block.
- **B.** \(W_1 < W_2\) because the vertical component of the applied force in Trial 2 reduces the normal force, allowing the block to move more easily.
- **C.** \(W_1 = W_2\) because work is a scalar quantity and is determined only by the magnitudes of the force and the displacement.
- **D.** \(W_1 = W_2\) because the same constant force magnitude is applied over the same displacement in both trials.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/113064/*