Question

Hilda holds a gardening book of weight \(10 \mathrm{N}\) at a height of $1.0 \mathrm{m}\( above her patio for \)50 \mathrm{s}$. How much work does she do on the book during that 50 s?

Short answer

Answer: Hilda does 0 J (joules) of work on the book during the 50 seconds.

Step-by-step solution

Identify the force and displacement of the book.

Hilda is holding the book steady, so the force she is applying on the book is equal to the weight of the book which is 10 N. Since the book is not displaced, the displacement is 0 m.

Calculate the work done by Hilda on the book.

The formula to calculate the work done is: \(W = F \cdot d \cdot \cos(\theta)\) where \(W\) is the work done, \(F\) is the force applied on the object, \(d\) is the displacement, and \(\theta\) is the angle between the force and the direction of displacement. In our case, \(F = 10\,\text{N}\), \(d = 0\,\text{m}\), and the force is applied in the same direction as the displacement, so \(\theta = 0^\circ\). However, there's no need to calculate the cosine of the angle, as the displacement is 0 m, so the work done by Hilda is: \(W = F \cdot d = 10\,\text{N} \cdot 0\,\text{m} = 0\,\text{J}\) So, Hilda does 0 J (joules) of work on the book during the 50 s.