Acceleration as a Function of Distance

AI Thread Summary
The discussion focuses on a physics problem involving a block on a frictionless incline being pulled by a force T. Key points include calculating work done by T, potential energy as a function of position, and deriving kinetic energy. The main challenge highlighted is defining acceleration as a function of position, with suggestions to apply Newton's Second Law and utilize the chain rule. Participants emphasize the need to consider all forces acting on the block, not just T. The conversation aims to clarify the approach to solving the problem effectively.
bollocks748
Messages
10
Reaction score
0

Homework Statement


Consider a block of mass M that is pulled up an incline by a force T that is parallel to the surface of the incline. The block starts from rest and is pulled a distance x by the force T. The incline, which is frictionless, makes an angle theta with respect to the horizontal.

1.1 Write down the work done by the force T.


1.2 Calculate the potential energy of the block as a function of the position x. From the work and the potential energy calculate the kinetic energy of the block as a function of position x.


1.3 Calculate the acceleration, a, of the block as a function of position x. Calculate the velocity of the block from the acceleration and hence the kinetic energy as a function of position. Show that the kinetic energies calculated in these two ways are equivalent.


1.4 Calculate the ratio of the potential to kinetic energy. What happens to this ratio for large T? Check your answer for the case theta = pi/2 which you should be able to recalculate easily.


Homework Equations



I didn't have problems with the first two parts, but I might as well put them here just in case:

Work done by T = T*x

PE(x)= m*g*x*sin(theta)

The Attempt at a Solution



What I'm really having trouble with is defining acceleration as a function of x, in part 3. I've had to do it before using the chain rule, but most of the time I was given a function of v(x), and integrated it, or used dv/dx and dx/dt to find dv/dt, or something of that nature. But I've never encountered a problem where I wasn't given some starter formula to work with... and I'm really at a loss here. Please help!
 
Physics news on Phys.org
bollocks748 said:
What I'm really having trouble with is defining acceleration as a function of x, in part 3. I've had to do it before using the chain rule, but most of the time I was given a function of v(x), and integrated it, or used dv/dx and dx/dt to find dv/dt, or something of that nature. But I've never encountered a problem where I wasn't given some starter formula to work with... and I'm really at a loss here. Please help!
You're on the right lines with use of the chain rule. Note that although you haven't been given an equation explicitly, you can formulate one yourself. Consider applying Newton's Second Law to the block.

Be aware that T is not the only force acting on the block.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top