Posted by winterWX on Monday, March 19, 2007 at 11:54pm.
A small block of mass m = 4.0 kg can slide along the frictionless loop-the-loop. The block is released from rest at point P, at height h = 23R above the bottom of the loop (R is the radius of the loop). Express your answers in the form ngR, where n is the number you will calculate.
a) What is the potential energy when the block is at point P? [at the very top of loop] (Assume the gravitational potential energy of the block Earth system is taken to be zero at the bottom of the loop).
b) What is the potential energy when the block is at point Q (halfway up the loop)?
c) What is the kinetic energy when the block is at the top of the loop?
For Further Reading
* Physics - drwls? - drwls, Tuesday, March 20, 2007 at 2:19am
We don't do homework for you here. Show us some effort and someone will critique your work or thought process.
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*note: i cant post the diagram.. but it looks like the number "6" slightly tilted 45 degrees counterclockwise
- potential energy for problem (a).. would be:
U = (4.0 kg)(9.8 m/s^2)(23R)
U = 901.6R
- problem (b) is similar:
U = (4.0 kg)(9.8 m/s^2)(2R)
- problem (c) would be:
K = 1/2mv^2
K = 1/2(4.0 kg)(???^2)
.. but what would i use if a velocity is not given?
----- and i'm also having difficulty understanding on how to express my answer in the form "ngR" .. what does that mean? .. is n just a random variable?
change in PE = - (change in KE). "V" is what you solve for.
I have no idea what they mean by "nGR"
On b, PE= mgh. If the block is halfway up the loop, wouldn't that height be R, not 2R>
On c,
Wouldn't the KE be equal to the change of PE, which would be the same as the original PE in the a)?
Your answer will be in the form of Number*gR . n the number is most certainly not random. It is what you calculate.
Winter:
A couple of notes for you. Please don't indentify tutors in the subject line. Most of us don't like our names in headlines. Don't do that.
Secondly, in reading your work, I think you would greatly benefit from a tutor once or twice a week. Discuss this with your parents. Another student I think could help you a lot. You don't seem to me to have a lot of confidence in your ability to think out these problems, and that is the crux of your difficulty. My experience tells me that a tutor would help greatly. My experience also tells me that it would be unwise to delay this decision.
To calculate the potential energy at point P (the top of the loop), you can use the formula:
U = mgh,
where U is the potential energy, m is the mass of the block (4.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height above the bottom of the loop (23R).
Substituting the values, we get:
U = (4.0 kg)(9.8 m/s^2)(23R) = 901.6R.
For point Q (halfway up the loop), the height is R, not 2R as stated in your post. So the potential energy, using the same formula, would be:
U = (4.0 kg)(9.8 m/s^2)(R) = 39.2R.
To calculate the kinetic energy at the top of the loop, we can use the principle that the change in potential energy is equal to the change in kinetic energy:
Change in PE = - Change in KE.
Since the block is released from rest, its initial kinetic energy is zero. Therefore, the change in kinetic energy is equal to the initial potential energy at point P. So the kinetic energy at the top of the loop is also 901.6R.
Regarding the expression "ngR," here's what it means:
- n represents a number that you will calculate based on the given values.
- g is the acceleration due to gravity (9.8 m/s^2).
- R is the radius of the loop, which is a given value.
So when expressing your answers in the form "ngR," you will substitute the calculated value of n, the constant g (9.8 m/s^2), and the given value of R.