Solving a 3-Part Frictionless Problem

  • Thread starter Thread starter jasminwater
  • Start date Start date
  • Tags Tags
    Frictionless
AI Thread Summary
The discussion revolves around solving a three-part physics problem involving a block sliding down a frictionless track. The block, starting from rest, travels a vertical distance of 1.9 meters and leaves the track horizontally, landing 4.61 meters away. The speed of the block upon leaving the track was calculated to be 7.407 m/s, and the height from which it was released was determined to be approximately 2.7963 meters using energy conservation principles. For the total speed upon impact, the user found the total energy to be 26.2143, leading to a calculated velocity of 9.599 m/s, but there was uncertainty about accounting for friction in this calculation. The discussion highlights the importance of applying kinetic and potential energy concepts while addressing the effects of friction in the problem.
jasminwater
Messages
3
Reaction score
0
I am stuck on this 3 part problem and don't really know what to dooo, any help would be greatly appreciated =)

A block (mass 569 grams) starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track. It leaves the track horizontally, flies through the air, and subsequently strikes the ground 4.61 m away. THe acceleration of gravity is 9.81 m/s^2. A) At what height h above the ground is the block released? B) What is the speed of the block when it leaves the track? C) What is the total speed of the block when it hits the ground?

The vertical distance between the horizontal section of the track and the ground is 1.9m. The height from where the block is released is uknown. And at the horizontal of the track there is a 1 m horizontal section where friction exists, and the coefficient of friction there is .3.


help please! thanksss =)
 
Physics news on Phys.org
Start by: Finding the speed as the block leaves the horizontal ramp. Use kinematics to find how long it would take an object to accelerate 1.9 m, distance from the track to the ground, then use that time and the distance 4.61 m to find the block's speed as it leaves the ramp. Hope that helps a lil.
 
Alright i did that, i found the speed as it leaves the track and i got : 7.407 m/s.

So to find the height... because KE = PE... then (mv^2)/2 = mgh --> h=(v^2)/2g and i got 2.7963 for my height...? is that correct?

Thats as far as I am getting, other than that i don't know how to do part c...

help again pleaaase
 
for part c, i found the total energy (KE + PE) and i got 26.2143. THen i did 26.2143 = (1/2)mv^2 and solved for v and got 9.599 m/s^2. Is that my velocity as it hits the ground? Dont i have to account for the friction?
 
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...

Similar threads

Back
Top