Solve Work-Energy Theorem Problems with Expert Help

[NguyenAdam]

I am currently working on a problem that I do not know how to solve. It is on the Work-Energy theorem. I would really appreciate it if you could help. :D

Kinetic Energy = 1/2 mv^2
Potential Energy = mgh
Work = Fd
KE(initial) + PE(initial) = KE(final) + PE(final)


The height above the ground of a child on a swing varies from 50 cm at the lowest point to 200 cm at the highest point. The maximum speed of the child is?

A. about 5.4 m/s
B. about 7.7 m/s
C. about 29 m/s
D. dependent on the child's mass

The answer for this question is A, and I do not understand how they got it.


I also have one more question.

The 2.0-kg head of an ax is moving at 60 m/s when it strikes the log. If the blade of the ax penetrates 20 mm into the log, the average force it exerts is

A. 3 kN
B. 90 kN
C. 72 kN
D. 180 kN

The answer is 180 kN.

I have no idea where to start for these problems. Help would really be appreciated!

 

[LawrenceC]

For the first question:

KE(initial) + PE(initial) = KE(final) + PE(final)

You have the correct equation. You only have to evaluate it. Don't forget about gravity.

 

[LawrenceC]

Hint for axe problem:

What energy does the axe use to cleave into the log? Does it do work on the log?

 

[tal444]

Hint for first question: Think of when the child is at his/her lowest point as 0 cm instead of 50 cm. What does that make his/her highest point equal to?

 

[asteriatic]

I would be happy to assist you with these problems. The first step in solving these problems is to understand the concepts of work, energy, and the work-energy theorem. Work is the force applied over a distance, and it can be calculated by multiplying the force by the distance traveled. Energy is the ability to do work, and it can be divided into two forms - kinetic energy, which is the energy of motion, and potential energy, which is the energy stored in an object due to its position or state. The work-energy theorem states that the total work done on an object is equal to the change in its kinetic and potential energy.

Now, let's apply these concepts to the first problem about the child on a swing. We know that at the lowest point, the child's potential energy is equal to mgh = 0.5 * m * 9.8 * 0.5 = 2.45 mJ. At the highest point, the potential energy is mgh = 0.5 * m * 9.8 * 2 = 9.8 mJ. We also know that at the highest point, all of the child's potential energy is converted into kinetic energy, which is calculated by KE = 0.5 * m * v^2. Therefore, we can set up the equation 2.45 mJ + 0.5 * m * v^2 = 9.8 mJ and solve for v to get the maximum speed of the child, which is approximately 5.4 m/s.

For the second problem, we can use the work-energy theorem again. The initial kinetic energy of the ax head is 0.5 * 2.0 kg * (60 m/s)^2 = 3.6 kJ. When it strikes the log, the kinetic energy is converted into work done on the log, which is equal to the force applied (unknown) multiplied by the distance traveled (0.02 m). Therefore, we can set up the equation 3.6 kJ = F * 0.02 m and solve for F to get approximately 180 kN.

I hope this explanation helps you understand how to approach and solve work-energy theorem problems. If you have any further questions, please do not hesitate to ask. Good luck with your studies!