Block and spring on a weird track (roller coaster)

In summary: Here I need to use that vf2 = vi2 + 2*a*d ?Ekf = 1/2*1.5kg*vf2Ekf = 1/2*1.5kg*(vf2 - vi2)Ekf = 1/2*1.5kg* (vf2 - 74.9999)Ekf = 1/2*1.5kg* (vf2 - 74.9999)Ekf = 1/2*1.5kg* (vf2 - 74.9999)Ekf = 1/2*1.5kg* (vf2 - 74.9999)The equations you have for
  • #1
masterchiefo
212
2

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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  • #2
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 ?
 
  • #3
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
 
  • #4
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 .
 
  • #5
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
 
  • #6
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
 
  • #7
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.
 
  • #8
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 :)
 
  • #9
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.
 

Related to Block and spring on a weird track (roller coaster)

1. What is a block and spring system on a roller coaster track?

A block and spring system on a roller coaster track is a mechanism that uses a combination of a block, which is a heavy object, and a spring, which is a flexible material, to generate kinetic energy and allow the roller coaster to move along the track.

2. How does a block and spring system work on a roller coaster track?

The block and spring system works by using the weight of the block to compress the spring. As the spring is released, it propels the block forward, generating kinetic energy that moves the roller coaster along the track. The block and spring are connected to the track through a series of pulleys and cables, allowing for a smooth and controlled movement.

3. What makes a roller coaster track "weird"?

A roller coaster track can be considered "weird" if it deviates from the traditional design of a straight or curved track. This can include twists, turns, loops, and other unconventional elements that add excitement and thrill to the ride.

4. How is the speed of a roller coaster with a block and spring system controlled?

The speed of a roller coaster with a block and spring system is controlled through the use of brakes and release mechanisms. The brakes can slow down or stop the block and spring, while the release mechanisms can control the amount of energy released from the spring, allowing for varying speeds throughout the ride.

5. Are there any safety concerns with using a block and spring system on a roller coaster track?

As with any roller coaster system, there are safety concerns that need to be addressed when using a block and spring system. These include proper maintenance and regular inspections to ensure the system is functioning correctly, as well as safety mechanisms in place to prevent accidents or malfunctions. It is important for engineers and designers to carefully consider all safety measures when implementing a block and spring system on a roller coaster track.

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