How much work is done by a tree to stop a 25.0 g arrow traveling at 75.0 m/s if it embeds itself 6.00 cm into the trunk?
Maybe Jared can answer your question ...
He did thank you
To calculate the work done by the tree to stop the arrow, we can use the work-energy principle. The work done by an object is equal to the change in its kinetic energy.
The kinetic energy of an object can be calculated using the formula: KE = (1/2) * m * v^2, where m is the mass of the object and v is its velocity.
In this case, the mass of the arrow is given as 25.0 g, which is equal to 0.025 kg, and the velocity is given as 75.0 m/s.
First, calculate the initial kinetic energy of the arrow before it embeds into the tree:
KE_initial = (1/2) * m * v_initial^2
= (1/2) * 0.025 kg * (75.0 m/s)^2
Next, calculate the final kinetic energy of the arrow after it embeds into the tree. Since the arrow comes to a stop, its final velocity is 0 m/s:
KE_final = (1/2) * m * v_final^2
= (1/2) * 0.025 kg * (0 m/s)^2
= 0 J
The work done by the tree to stop the arrow is the difference between the initial and final kinetic energies:
Work = KE_final - KE_initial
Work = 0 J - [(1/2) * 0.025 kg * (75.0 m/s)^2]
Now, let's calculate the work done by the tree:
Work = - (1/2) * 0.025 kg * (75.0 m/s)^2
Work = - 70.31 J
Therefore, the work done by the tree to stop the arrow is approximately -70.31 Joules. The negative sign indicates that the work done is in the opposite direction of motion, which implies that the tree absorbed the energy from the arrow.