# Can someone check my answer for this work-energy problem?

• toesockshoe
In summary, the conversation discussed finding the tension in a string when a mass at the end of the string swings up to its highest vertical point. Calculations were made using the equations for net work and force, and the effects of gravity and centripetal acceleration were considered. It was determined that the final answer for the tension force is -mg - (2mg*(-L+2*x))/(L-x). In the case where L=x, there will be no centripetal acceleration and the equation will need to be modified.
toesockshoe

## Homework Statement

A mass attached to a string of length L is released from rest from an initial horizontal position (although the diagram implies it is not that high, please ignore that part of the diargram - sorry). From veritcally below where the string is attached at its top, there is a peg a distance x below the top attachment of the string. As the mass swings down to its lowest position, the string then touches the peg and the mass swings up and around from this new point (i.e. the peg's position). The mass continues to swing upward. Find the tension in the string when the mass is at its highest vertical point.

The diagram is here: http://nebula.deanza.edu/~Newton/4A/4AHWSet6.html (it is number 5).

## Homework Equations

$$W_net=\Delta E$$
$$\vec{F} = m\vec{a}$$

## The Attempt at a Solution

I think the highest point is when it goes around the peg and is completely vertical. (I think I am supposed to disregard the position when the peg is just released from rest, in which that would be the highest point).

so I first found the velocity at the point:

System: Earth and Mass
[/B]
$$W_{net}=\Delta E$$
tension force does no work becuase it is normal to the displacement of the mass
so:
$$0 = \Delta GPE + \Delta KE$$
$$0 = -mg(2x-L) + \frac{1}{2}mv_f^2$$
2x-L is the change of vertical displacement from the position at rest and when it vertically at the highest point after going passed the peg. so:

$$v_f=\sqrt{2g(2x-L)}$$Now, I am trying to find the tension by doing an $$\vec{F}=m \vec{a}$$ problem

Let the system be the mass with the y direction going up and the x direction to the left:

$$F_t - F_g = ma_c$$
$$F_t = \frac{mv^2}{L-x}+mg$$
^the above step is where I am a bit unsure of myself... do I need to include gravity into my centripetal acceleration? if so? can I just say that the centripetal acceleration is as follows:

$$a_c=\frac{mv^2g}{L-x}$$ ?

ok so i continue and I plug in the velocity I found through the energy work above and plug it into the velocity variable in the F=ma solution to get:

$$F_t = \frac{2mg(2x-L)}{L-x} +mg$$

can someone check my work? I'm not sure if my answer makes sense because if L=x, then my Force value would go berserk...

I gather you are taking g as a positive quantity. Consider the mass at speed v at the top of the loop around the peg. Would a greater g increase or decrease the tension?

haruspex said:
I gather you are taking g as a positive quantity. Consider the mass at speed v at the top of the loop around the peg. Would a greater g increase or decrease the tension?
my apogies: i would like to change my F=ma coordinate system as y going downard and I would have:

Fg -Ft = ma

then my final answers would be:

$$mg - \frac{2mg(2x-L)}{L-x}$$

is that correct?

toesockshoe said:
my apogies: i would like to change my F=ma coordinate system as y going downard and I would have:

Fg -Ft = ma

then my final answers would be:

$$mg - \frac{2mg(2x-L)}{L-x}$$

is that correct?
You still have a sign wrong somewhere.
What are the forces on the mass at the point of interest? Which way does each act? Which way does their resultant act?

I got
$$F_T = -mg - \frac{2mg*(-L+2*x)}{L-x}$$

As for what happens when L=x.. well... I'm not sure there will be any tension force using the equation you got above with ac, since there will be no centripetal acceleration and the whole thing doesn't make sense (the string isn't long enough to go around the peg: no centripetal acceleration, so your F=ma equation will need to be modified).

See you in class tomorrow! Only a few more weeks then Exam 2, and Final. I will give Newton a nice shiny red apple on exam day.

Last edited:
yawaraf said:
I got
$$F_T = -mg - \frac{2mg*(-L+2*x)}{L-x}$$

As for what happens when L=x.. well... I'm not sure there will be any tension force using the equation you got above with ac, since there will be no centripetal acceleration and the whole thing doesn't make sense (the string isn't long enough to go around the peg: no centripetal acceleration, so your F=ma equation will need to be modified).

See you in class tomorrow! Only a few more weeks then Exam 2, and Final. I will give Newton a nice shiny red apple on exam day.
haha cool. see ya!

## 1. What is a work-energy problem?

A work-energy problem is a type of physics problem that involves calculating the work done by an object or force on another object. It typically involves the use of the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

## 2. Why do I need someone to check my answer for a work-energy problem?

Work-energy problems can be complex and involve multiple steps and calculations. Having someone else check your answer can help ensure that you did not make any mistakes or overlook any important details.

## 3. What is the most common mistake made in work-energy problems?

The most common mistake in work-energy problems is incorrectly calculating the work done by a force. This can happen if the direction of the force is not taken into account or if the magnitude of the force is not properly calculated.

## 4. Is it important to show all of my work when solving a work-energy problem?

Yes, it is important to show all of your work when solving a work-energy problem. This allows you to track your steps and identify any errors, and it also helps others understand your thought process and check your work.

## 5. Can someone check my answer for a work-energy problem even if they are not an expert in physics?

Yes, someone without expertise in physics can still check your answer for a work-energy problem. While they may not be able to provide insights into the underlying concepts, they can still check your calculations and point out any potential errors.

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