Two carts connected with a spring -- working with forces and acceleration

In summary, to pull the 8kg cart with the same amount of force as pulling the 2kg cart, you would need to exert a force of 150N.
  • #1
Jaccobtw
163
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Homework Statement
A 2.0 kg cart and an 8 kg cart are connected by a relaxed, horizontal spring of spring constant 300 N/m. You pull the 8 kg cart with some constant horizontal force. The separation between the carts increases for a short time interval, then remains constant as you continue to pull and the spring is stretched by 0.1 m. (a) What pulling force did you exert? (b) If you instead pull the 2.0 kg cart, what forces must you exert to get the same stretch in the spring?
Relevant Equations
F = -kd
F = ma
Problem Statement: A 2.0 kg cart and an 8 kg cart are connected by a relaxed, horizontal spring of spring constant 300 N/m. You pull the 8 kg cart with some constant horizontal force. The separation between the carts increases for a short time interval, then remains constant as you continue to pull and the spring is stretched by 0.1 m. (a) What pulling force did you exert? (b) If you instead pull the 2.0 kg cart, what forces must you exert to get the same stretch in the spring?
Relevant Equations: F = -kd
F = ma

I'm not sure how to execute the problem.
 
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  • #2
How much force is the spring exerting once it has been stretched?
 
  • #3
Welcome to the PF. :smile:
Jaccobtw said:
I'm not sure how to execute the problem.
You need to show some effort before we can be of tutorial help. Please use the Relevant Equations to set up the equations for both cases. Draw the Free Body Diagrams (FBDs) for each case, and show all of the forces involved. Please show us your work...
 
  • #4
How do you draw a freebody diagram on here?
 
  • #5
.Scott said:
How much force is the spring exerting once it has been stretched?
F = -kd
(-300)(-0.1) = 30 N
 
  • #6
Jaccobtw said:
How do you draw a freebody diagram on here?
We don't have any built-in drawing tools (yet), but you can make a hand sketch and attache it to a post with the Attach Files button below the Edit window. Try to take a good quality picture of the sketch, or better yet scan it with a scanner/copier.
 
  • #7
.Scott said:
How much force is the spring exerting once it has been stretched?
Ok, here is my understanding: Because (technically) the only thing exerting a force on the 2 kg car is the spring we can do this:

-kd = ma

and solve for acceleration (15m/s^2)

Now, because the 2kg cart and the 8 kilogram cart are a single moving system, the 8 kg cart must also be accelerating at the same rate. Also, we still have the spring pulling in the opposite direction as the car:

F = ma - kd
(8kg)(15m/s) - (300)(0.1)

F = 1.5 x 10^2 N

but technically this doesn't make sense because it means the cart was never accelerating at 15m/s^2 because the spring was pulling against it. I don't know, this is why I'm confused
 
  • #8
Jaccobtw said:
Now, because the 2kg cart and the 8 kilogram cart are a single moving system ...
And this system is how much more massive that the cart being pulled at 30N?
 
  • #9
.Scott said:
And this system is how much more massive that the cart being pulled at 30N?
8N right?
 
  • #10
Jaccobtw said:
8N right?
Let's try that again:
This system is how much more massive that the cart being pulled at 30N?
That is, what is the ratio of the total mass of both carts to the mass of the one being pulled by 30N?
 
  • #11
.Scott said:
Let's try that again:
This system is how much more massive that the cart being pulled at 30N?
That is, what is the ratio of the total mass of both carts to the mass of the one being pulled by 30N?
Sorry, I meant 6 kg (8kg - 2kg) = 6kg. The ratio is 5kg to 1 kg (8 + 2 / 2)
 
  • #12
Jaccobtw said:
Sorry, I meant 6 kg (8kg - 2kg) = 6kg. The ratio is 5kg to 1 kg (8 + 2 / 2)
Can we call that a factor of five?
Think - what force will be required to pull that entire mass at the same rate as the smaller one?
 
  • #13
.Scott said:
Can we call that a factor of five?
Think - what force will be required to pull that entire mass at the same rate as the smaller one?
10kg x 15m/s^2 = 150 N
 
  • #14
Yup. It was 30N for the 2Kg cart - so it would be 5 times the force for 5 times the mass. 5x30N.
 
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Likes Jaccobtw
  • #15
Thank you so much...haha that went over my head at first
 

1. How does a spring affect the motion of two connected carts?

A spring affects the motion of two connected carts by exerting a force on each cart, causing them to move in opposite directions. When one cart moves, it pulls on the spring, which then exerts a force on the other cart, causing it to move in the opposite direction. This creates a back-and-forth motion between the two carts.

2. What is the relationship between the spring force and the displacement of the carts?

The spring force is directly proportional to the displacement of the carts. This means that the more the carts are displaced from their equilibrium position, the greater the force exerted by the spring.

3. How does the mass of the carts affect the motion?

The mass of the carts affects the motion by changing the acceleration of each cart. According to Newton's Second Law, a larger mass will require a greater force to accelerate, so the heavier cart will move slower compared to the lighter cart.

4. Can the spring force ever be greater than the weight of the carts?

Yes, the spring force can be greater than the weight of the carts. This can happen when the carts are displaced from their equilibrium position and the spring is stretched or compressed. In this case, the spring force can exceed the weight of the carts and cause them to accelerate in the direction of the stronger force.

5. What happens to the motion of the carts when the spring is removed?

When the spring is removed, the motion of the carts will change because there is no longer a force acting between them. The carts will continue to move at a constant velocity unless acted upon by another force, such as friction. Without the spring, there will be no back-and-forth motion between the carts.

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