A 2.22 kg book is dropped from a height of +3.2 m. What is its acceleration?
can you help?
Gravity is what makes if fall towards Earth. THe gavitational acceleration near the surface of Earth is 9.8m/s^2 . I recommend you commit that to memory.
its not 9.8 or 9.81, ive already tried that
Its Negative 9.81
Of course! To find the acceleration of the book, you can use the equation for gravitational potential energy. The equation is:
PE = mgh
Where:
PE is the gravitational potential energy,
m is the mass of the object,
g is the acceleration due to gravity, and
h is the height.
In this case, the mass of the book is 2.22 kg and the height it is dropped from is 3.2 m.
Now, since the book is dropped (meaning it is freely falling), all of its potential energy is converted into kinetic energy, according to the law of conservation of energy. The equation for kinetic energy is:
KE = (1/2)mv^2
Where:
KE is the kinetic energy,
m is the same mass as before,
and v is the velocity.
Since the book starts from rest (at the top of the fall), its initial velocity is 0. So we can simplify the equation to:
KE = (1/2)mv^2 = (1/2)m(0)^2 = 0
Therefore, all of the gravitational potential energy is converted into kinetic energy, and KE = PE.
Now, let's set up the equation:
mgh = (1/2)mv^2
The mass cancels out from both sides of the equation:
gh = (1/2)v^2
Now, divide both sides of the equation by h:
g = (1/2)v^2 / h
Since we know h = 3.2 m, and the initial velocity is 0, we can simplify further:
g = (1/2) * (0)^2 / 3.2
g = 0 / 3.2
g = 0
Therefore, the acceleration of the book is 0 m/s². This means that the book is not accelerating, but rather experiencing free-fall due to gravity.