Question

1. a horizontal force of 20 newtons eastward causes a 10 - kilogram box to have a displacement of 5 meters eastward. what is the total work done on the box by the 20 - newton force?

Explanation:

Step1: Recall the work formula

The formula for work \( W \) is \( W = F \cdot d \cdot \cos\theta \), where \( F \) is the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement vectors.

Step2: Determine the angle

Since the force and displacement are in the same direction (eastward), \( \theta = 0^\circ \), and \( \cos(0^\circ)=1 \).

Step3: Substitute values into the formula

Given \( F = 20\,\text{N} \), \( d = 5\,\text{m} \), and \( \cos\theta = 1 \), we substitute into \( W = F \cdot d \cdot \cos\theta \):
\( W = 20\,\text{N} \times 5\,\text{m} \times 1 \)

Step4: Calculate the work

\( 20\times5 = 100 \), so \( W = 100\,\text{J} \) (joules, since \( 1\,\text{J} = 1\,\text{N}\cdot\text{m} \)).

Answer:

The total work done is \( 100 \) joules.