How Fast Does a Book Fall from 4.20 Meters?

[kissafilipino]

A physics book of unknown mass is dropped 4.20 m. What speed does the book have just before it hits the ground? Unless otherwise directed, assume that air resistance is negligible.

Height: 4.20
Acceleration: 9.81
Mass: ?
Velocity: ?

The answer box is for velocity: ___ m/s

Homework Equations

1)Kinetic Energy: 1/2*m*v^2
2)Kinetic Energy: (p^2)/2*m

3)Potential Grav. Energy: m*g*(delta or change in)h

The Attempt at a Solution

I thought the whole point was to look at how the Potential energy becomes kinetic energy, and so I went about it by using the Potential energy formula. m(9.81)(4.40) = 43.164 however I don't know if I that gives me mass or anything, I am stumped because I think I do not have enough information to answer this question. Plus going about it by using another formula but that does not seem to be working either, such as the Force = mass times acceleration, again I do not know mass, or the power = force times velocity, again I need to find force, which I do not know. Please help me!

 

[MillerGenuine]

U= potential energy
K= kinetic energy
Conservation of energy => Kf + Uf = Ki + Ui
What is the initial Kinetic Energy? (it has v=0)
What is the final potential energy? (it has h=0)
this should simplify this equation so that we have our kinetic = potential
And when you solve..the mass should cancel out.

 

[a.ratnaparkhi]

I'm satisfied with MillerGenuine's method, But I have another method. In the problem you know initial velocity which is zero, total displacement i.e. height, acceleration.
So you can think of
vf^2=vi^2+2as.

 

[kissafilipino]

THANKYOU, both your equations and formulas got the same answer (of course) but now I understand how it is possible and two ways of solving this type of question.

 

[rwisz]

Hello,

Thank you for reaching out for help with this problem. I can see that you have made some attempts at solving it, so I will guide you through the correct approach.

First, let's review the conservation of energy principle. This principle states that energy cannot be created or destroyed, it can only be transferred from one form to another. In this case, we are looking at the potential energy of the book (due to its position above the ground) being converted into kinetic energy (due to its motion).

Now, let's look at the equations you have listed. The first two equations are both for kinetic energy, but they are slightly different. The first one is the standard equation for kinetic energy, where m is the mass of the object and v is its velocity. The second equation is the relativistic equation for kinetic energy, where p is the momentum of the object and m is its mass. Since we are dealing with a non-relativistic situation, we will stick to the first equation.

The third equation is for potential gravitational energy, which is the energy an object has due to its position in a gravitational field. This equation is mgh, where m is the mass of the object, g is the acceleration due to gravity (9.81 m/s^2 on Earth), and h is the height of the object.

Now, let's apply these equations to the problem. We know that the book is dropped from a height of 4.20 m, so we can plug that value into the third equation and set it equal to the first equation, since the potential energy is being converted into kinetic energy. This gives us:

mgh = 1/2*m*v^2

We can then rearrange this equation to solve for v:

v = √(2gh)

Now, we need to find the value of g. We are told that air resistance is negligible, so we can assume that the book is in free fall and g = 9.81 m/s^2. We can then plug this value into the equation and solve for v:

v = √(2*9.81*4.20) = 9.79 m/s

Therefore, the book has a speed of 9.79 m/s just before it hits the ground.

I hope this explanation helps you understand the problem and how to approach it. Remember to always consider the conservation of energy principle and use the appropriate equations for the situation