Applying the work-energy theorem

[hpnhpluv]

Homework Statement

Use the work-energy theorem to solve.
A branch falls from the top of a 95 m tall tree, starting from rest. How fast is it moving when it reaches the ground? Neglect air resistance.

Homework Equations

work-energy theorem:w_total=K_2-K_1
In this problem, K_1 is 0 since it is starting from rest.
K_2=(1/2)*m*v^2 (?)
w=fd (relevant?)

The Attempt at a Solution

If I use the equations I have above, I end up with w=1/2*m*v_f^2.
However, I am not given a mass. I do not know how to finish the calculation without a mass.
If I use w=fd, I have (9.8 m/s^2)(mass)*(95 m).
I am really stuck.

 

[atyy]

In addition to the equations you listed, you also need the equation for the force of gravity on a mass.

You are only asked for the final velocity. Assume a mass, and carry through with the equations. Hopefully the mass will cancel out when you solve for the velocity.

 

[baba89]

I would like to clarify that the work-energy theorem states that the total work done on an object is equal to the change in its kinetic energy. In this problem, the work done on the branch is due to the force of gravity, and the change in its kinetic energy is equal to its final kinetic energy.

To solve this problem, we can use the equation for gravitational potential energy, U=mgh, where m is the mass of the branch, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the tree (95 m). Since the branch starts from rest, its initial kinetic energy is 0. Therefore, the total work done on the branch is equal to its change in potential energy, or W=U_f-U_i=mgh.

Since we do not know the mass of the branch, we cannot solve for the final velocity using the work-energy theorem. However, we can use the equation for velocity in terms of potential energy, v=sqrt(2gh), to find the final velocity. Assuming the branch falls straight down, the potential energy at the top of the tree (U_i) is equal to the kinetic energy at the bottom (K_f), so we can write the equation as mgh=1/2mv_f^2. Solving for v_f, we get v_f=sqrt(2gh).

Substituting the given values, we get v_f=sqrt(2(9.8 m/s^2)(95 m))=43.06 m/s.

Therefore, the branch will be moving at a speed of 43.06 m/s when it reaches the ground. It is important to note that this calculation neglects the effects of air resistance, which would likely decrease the final velocity of the branch.