You drop a 1.60 kg book to a friend who stands on the ground at distance D = 12.0 m below. If your friend's outstretched hands are at distance d = 1.40 m above the ground (see the figure), (a) how much work Wg does the gravitational force do on the book as it drops to her hands? (b) What is the change ΔU in the gravitational potential energy of the book-Earth system during the drop?
To solve this problem, we need to use the formulas for work (W) and gravitational potential energy (U) and substitute the given values.
(a) To find the work done by the gravitational force, we use the formula:
W = m * g * h
where:
m = mass of the book = 1.60 kg
g = acceleration due to gravity = 9.8 m/s^2 (approximate value on Earth)
h = height through which the book falls = D - d = 12.0 m - 1.40 m = 10.6 m
Substituting the values into the formula, we get:
W = 1.60 kg * 9.8 m/s^2 * 10.6 m
W = 168.448 J (rounded to three significant figures)
Therefore, the gravitational force does approximately 168.448 J of work on the book.
(b) To find the change in gravitational potential energy, we use the formula:
ΔU = m * g * Δh
where:
ΔU = change in gravitational potential energy
m = mass of the book = 1.60 kg
g = acceleration due to gravity = 9.8 m/s^2 (approximate value on Earth)
Δh = change in height = h = D - d = 12.0 m - 1.40 m = 10.6 m
Substituting the values into the formula, we get:
ΔU = 1.60 kg * 9.8 m/s^2 * 10.6 m
ΔU = 165.664 J (rounded to three significant figures)
Therefore, the change in gravitational potential energy of the book-Earth system during the drop is approximately 165.664 J.
To find the work done by the gravitational force on the book as it drops to your friend's hands (a), we can use the formula:
Work (W) = Force (F) x Distance (d) x cosine(angle)
In this case, the force acting on the book is the gravitational force given by:
Force (F) = mass (m) x acceleration due to gravity (g)
Where:
mass (m) = 1.60 kg (given)
acceleration due to gravity (g) = 9.8 m/s^2 (standard value on Earth)
Substituting these values into the formula, we have:
Force (F) = 1.60 kg x 9.8 m/s^2 = 15.68 N
Now, we need to find the angle between the direction of the gravitational force and the direction of displacement, which is the angle at which the book is dropped. Since we don't have this information in the question, we assume it to be 0 degrees for simplicity.
Thus, cosine(angle) = cos(0°) = 1
Now, we can calculate the work done by the gravitational force:
Work (Wg) = Force (F) x Distance (d) x cosine(angle)
Wg = 15.68 N x 12.0 m x 1 = 188.16 J (Joules)
Therefore, the work done by the gravitational force on the book as it drops to your friend's hands is 188.16 Joules (a).
To find the change in gravitational potential energy of the book-Earth system during the drop (b), we can use the formula:
Change in Gravitational Potential Energy (ΔU) = m x g x Δh
In this case, the change in height (Δh) is given by:
Δh = D - d = 12.0 m - 1.40 m = 10.6 m
Now, we can substitute the known values into the formula:
ΔU = mass (m) x acceleration due to gravity (g) x change in height (Δh)
ΔU = 1.60 kg x 9.8 m/s^2 x 10.6 m = 168.448 J (Joules)
Therefore, the change in the gravitational potential energy of the book-Earth system during the drop is 168.448 Joules (b).
1561
work=force*distance =mg*(12-1.4)
change PE=mgh