You drop a 2.00 kg book to a friend who stands on the ground at distance D= 10.0 m below. If your friend’s outstretched hands are at distance d=1.50 m above the ground see the next figure, (a) how much work W_g does the gravitational force do on the book as it drops to her hands? (b) What is the change ∆U in the gravitational potential energy of the book–Earth system during the drop? If the gravitational potential energy U of that system is taken to be zero at ground level, what is U (c) when the book is released and (d) when it reaches her hands?
q11
To find the answers to these questions, we need to use the concept of work and gravitational potential energy. Let's break it down step by step:
(a) To calculate the work done by the gravitational force on the book as it drops to your friend's hands, we use the equation W_g = m * g * h, where m is the mass of the book, g is the acceleration due to gravity, and h is the vertical distance.
In this case, the mass of the book is given as 2.00 kg, the acceleration due to gravity can be taken as 9.8 m/s^2, and the vertical distance is the sum of the height of your friend's hands (1.50 m) and the distance below the ground (10.0 m). So, the total distance is 11.50 m.
Plugging in these values, we get W_g = 2.00 kg * 9.8 m/s^2 * 11.50 m ≈ 225.40 J.
Therefore, the gravitational force does about 225.40 Joules of work on the book as it drops to your friend's hands.
(b) The change in gravitational potential energy (∆U) of the book-Earth system during the drop can be calculated using the equation ∆U = m * g * ∆h, where ∆h is the change in height.
In this case, the initial height is 0 (ground level). So, the change in height (∆h) is the final height which is 1.50 m.
Plugging in the values, we get ∆U = 2.00 kg * 9.8 m/s^2 * 1.50 m ≈ 29.40 J.
Therefore, the change in gravitational potential energy of the book-Earth system during the drop is approximately 29.40 Joules.
(c) When the book is released, the gravitational potential energy (U) of the book-Earth system is taken to be zero at ground level. So, U = 0 Joules.
(d) When the book reaches your friend's hands, we can again use the equation U = m * g * h. As mentioned earlier, the height is the sum of your friend's hand height (1.50 m) and the distance below the ground (10.0 m). So, the total height is 11.50 m.
Plugging in the values, we get U = 2.00 kg * 9.8 m/s^2 * 11.50 m ≈ 225.40 J.
Therefore, when the book reaches your friend's hands, the gravitational potential energy of the book-Earth system is approximately 225.40 Joules.
In summary:
(a) The gravitational force does about 225.40 Joules of work on the book.
(b) The change in gravitational potential energy (∆U) of the book-Earth system is approximately 29.40 Joules.
(c) When the book is released, the gravitational potential energy (U) is 0 Joules.
(d) When the book reaches your friend's hands, the gravitational potential energy (U) is approximately 225.40 Joules.