Block and spring on a weird track (roller coaster)

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Homework Help Overview

The problem involves a block of mass 1.5 kg launched from a point on a curved track with an initial speed of 12 m/s. The objective is to determine the speed of the block at point C, as well as other related quantities such as centripetal acceleration and normal force at that point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of conservation of energy to find the speed at point C and raise questions about the initial kinetic energy and potential energy changes as the block moves along the track.
  • Some participants express confusion about the effects of the curved ramp and circular motion on the calculations.
  • There are inquiries regarding the direction of forces acting on the block and the implications of centripetal acceleration on the normal force.

Discussion Status

Participants are actively engaging with the problem, sharing calculations for kinetic and potential energy, and discussing the implications of their results. Some guidance has been provided regarding the correct application of energy principles and the interpretation of forces acting on the block, though there is still uncertainty about certain aspects of the problem.

Contextual Notes

Participants note the complexity introduced by the curved track and the circular motion, which leads to questions about assumptions regarding forces and accelerations. There is also mention of the lack of explicit consensus on the correct interpretation of the normal force and its direction.

masterchiefo
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Homework Statement


A block of mass m = 1.5 kg is launched from point A with speed VA = 12 m / s, in the circuit
shown in Figure 4. If we assume that there is no friction determine:
a) the speed of the block when it reaches the point C;

Drawing of the problem:
http://i.imgur.com/TrcMJ5I.png

Homework Equations


This question has about 10 questions but I only want to really understand the first one and I would like your help with this because its very confusing me since the ramp is curved and not straight plane like usual.
Also the circular part is confusing me a lot too.

The Attempt at a Solution


Do I have to use the conservative energy formula to find the final speed to C ?

I am very sorry I don't have anything as my own work (try) but I really do not know how to do this.
I have never posted a help thread before without any work done.
 
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masterchiefo said:

Homework Statement


A block of mass m = 1.5 kg is launched from point A with speed VA = 12 m / s, in the circuit
shown in Figure 4. If we assume that there is no friction determine:
a) the speed of the block when it reaches the point C;

Drawing of the problem:
http://i.imgur.com/TrcMJ5I.png

Homework Equations


This question has about 10 questions but I only want to really understand the first one and I would like your help with this because its very confusing me since the ramp is curved and not straight plane like usual.
Also the circular part is confusing me a lot too.

The Attempt at a Solution


Do I have to use the conservative energy formula to find the final speed to C ?

I am very sorry I don't have anything as my own work (try) but I really do not know how to do this.
I have never posted a help thread before without any work done.
TrcMJ5I.png


What is the initial Kinetic Energy of the block?

What is the change in potential energy of the block in going from point A to point C ?
 
SammyS said:
TrcMJ5I.png


What is the initial Kinetic Energy of the block?

What is the change in potential energy of the block in going from point A to point C ?
Ek = 1/2*m*vi2
Ek = 1/2*1.5kg*12m/s
Ek = 9J

Ugf-Ugi = m*g*hf - m*g*hi
(1.5kg*9.81*0) - 1.5kg*9.81*(2m+1.5m)
= -51.5025J
 
masterchiefo said:
Ek = 1/2*m*vi2
Ek = 1/2*1.5kg*12m/s
Ek = 9J

Ugf-Ugi = m*g*hf - m*g*hi
(1.5kg*9.81*0) - 1.5kg*9.81*(2m+1.5m)
= -51.5025J
You didn't square the velocity for kinetic energy.

You have the potential energies reversed. UC - UA is positive .
 
SammyS said:
You didn't square the velocity for kinetic energy.

You have the potential energies reversed. UC - UA is positive .
sorry my bad.
Ek= 108J
and potential energy is 51.5025J

Then I use this formula to find the Vf?
Ugi+1/2*m*vi2 = Ugf + 1/2*m*vf2
108J = 51.5025J+ 1/2*1.5Kg*vf2
vf = 8.67929m/s
 
SammyS said:
You didn't square the velocity for kinetic energy.

You have the potential energies reversed. UC - UA is positive .
Then
b) the centripetal acceleration ( normal) in point C;
I would simply do this?
ac = v2/R
ac = 8.679292/1.5
=50.22m/s2

c) the normal reaction exerted by the loop on the block at point C (Draw the diagram of the forces and acceleration) ;
Do not really understand this.

Would N be the same direction as W in this situation since the block is under the plane.
Forces in Y = 0 = - N - W
= 0 = - N - 1.5*9.81
N = -14.715N
 
masterchiefo said:
Would N be the same direction as W in this situation since the block is under the plan
Yes, but the following is not correct. The acceleration is not zero. See part (b).
masterchiefo said:
Forces in Y = 0 = - N - W
= 0 = - N - 1.5*9.81
N = -14.715N
You assumed the normal force was downward, but the result came out negative, which would make the normal force be upward.
 
SammyS said:
Yes, but the following is not correct. The acceleration is not zero. See part (b).

You assumed the normal force was downward, but the result came out negative, which would make the normal force be upward.
Then the gravity is upward as well?
SammyS said:
Yes, but the following is not correct. The acceleration is not zero. See part (b).

You assumed the normal force was downward, but the result came out negative, which would make the normal force be upward.
http://i.imgur.com/jzuoGuC.png is my diagram of forces and acceleration correct?

If that is correct then I can do this:
Forces in Y = - m*ac = - N - W
Forces in Y = - (1.5kg*50.22m/s2c) = - N - (1.5kg*9,81)
now my N is 60.615N

is that correct?

Thanks for helping me, I really appreciate it :)
 
In reply to a comment of mine you wrote:
masterchiefo said:
Then the gravity is upward as well?

No. Gravity is downward. The fact that N turns out to be upward came from you ignoring centripetal acceleration. If the block were to remain stationary on the bottom side of the track at point C, then the track would have to exert a force upward to compensate for gravity.
http://i.imgur.com/jzuoGuC.png is my diagram of forces and acceleration correct?

If that is correct then I can do this:
Forces in Y = - m*ac = - N - W
Forces in Y = - (1.5kg*50.22m/s2c) = - N - (1.5kg*9,81)
now my N is 60.615N

is that correct?

Thanks for helping me, I really appreciate it :)
That is a correct method. Looks good.
 
  • #10
SammyS said:
In reply to a comment of mine you wrote:No. Gravity is downward. The fact that N turns out to be upward came from you ignoring centripetal acceleration. If the block were to remain stationary on the bottom side of the track at point C, then the track would have to exert a force upward to compensate for gravity.

That is a correct method. Looks good.
Now to find :
d) the speed of the block as it passes under D and E;

For D:

Ugi = m*g*hi = 1.5kg*9.81*(2+1.5) = 51.5025J
Ugf = m*g*hf = 1.5kg*9.81*(2) = 29.43J
Eki = 1/2*m*vi2 = 1/2*1.5kg*8.679292 =56.4976J
Ekf = 1/2*m*vf2 = 1/2*1.5kg*vf2

Ugi + Eki = Ugf +Ekf
51.5025J + 56.4976J= 29.43J + 1/2*1.5kg*vf2
vf = 10.2352m/s ---> D

and now for E;

Ugi = m*g*hi = 1.5kg*9.81*(2) = 29.43J
Ugf = m*g*hf = 1.5kg*9.81*(2) = 29.43J
Eki = 1/2*m*vi2 = 1/2*1.5kg*10.2352m/s2 =78.5695J
Ekf = 1/2*m*vf2 = 1/2*1.5kg*vf2

Ugf - Ugi = 0
Eki = Ekf
78.5695J = 1/2*1.5kg*vf2
= 10.2342m/s

normal that my speed is the same ? there is no acceleration I should add somewhere :/?

thanks.
 

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