Question

question 13 (1 point)
you lift a 10-n physics book up in the air a distance of 1.0 m, at a constant velocity of 0.50 m/s. what is the work done by the weight of the book?
+10 j
-10 j
+5.0 j
-5.0 j

Explanation:

Step1: Recall the work formula

The formula for work 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: Identify force, displacement, and angle

The weight of the book is \( F = 10\,\text{N} \) (acting downward). The displacement \( d = 1.0\,\text{m} \) is upward. The angle between the weight (downward) and displacement (upward) is \( \theta = 180^\circ \), and \( \cos(180^\circ) = -1 \).

Step3: Calculate work

Substitute into the formula: \( W = 10\,\text{N} \times 1.0\,\text{m} \times \cos(180^\circ) = 10 \times 1 \times (-1) = -10\,\text{J} \).

Answer:

-10 J (corresponding to the option: -10 J)