# How to find the position update for Step 2 in Spring Problem

• Troi Jones
In summary: Earth...is greater than the upward force on the block by the spring.FEarth,y = -0.6867 NFnet,y = 1.6133 NIn summary, the block moves upward due to the greater upward force on the block by the spring.
Troi Jones

## Homework Statement

A spring has a relaxed length of 35 cm (0.35 m) and its spring stiffness is 10 N/m. You glue a 70 gram block (0.070 kg) to the top of the spring, and push the block down, compressing the spring so its total length is 12 cm. You make sure the block is at rest, then at time t = 0 you quickly move your hand away. The block begins to move upward, because the upward force on the block by the spring is greater than the downward force on the block by the Earth. Calculate approximately y vs. time for the block during a 0.27-second interval after you release the block, by applying the Momentum Principle in three steps each of 0.09-second duration.

## Homework Equations

Fspring= kx
Fgrav= ma
Fnet= Fspring + Fgrav
p= mv
delta p= F*delta t
update position= Vavg*delta t= delta r

## The Attempt at a Solution

STEP 1[/B]

Force: Just after releasing the block, calculate the force exerted on the block by the spring, the force exerted on the block by the Earth, and the net force:
Fspring,y = 2.3 NFEarth,y = -0.6867 N
Fnet,y = 1.6133 N

Momentum update: Just after releasing the block, the momentum of the block is zero. Approximate the average net force during the next time interval by the force you just calculated. At t = 0.09 seconds, what will the new momentum and velocity of the block be?
py = 0.145197 kg · m/s
vy = 2.074 m/s

Position update: Initially the bottom of the block is at y = 0.12 m. Approximating the average velocity in the first time interval by the final velocity, what will be the new position of the bottom of the block at time t = 0.09 seconds?
y = 0.30666 mSTEP 2

Force: At the new position, calculate the force exerted on the block by the spring, the force exerted on the block by the Earth, and the net force:
Fspring,y = 0.43 N
FEarth,y = -0.6867 N
Fnet,y = -0.2567 N

Momentum update: Approximate the average net force during the next time interval by the force you just calculated. At time t = 2 × 0.09 = 0.18 seconds, what will the new momentum and velocity of the block be?
py = 0.122094 kg · m/s
vy = 1.744 m/s

Position update: Approximating the average velocity in the second time interval by the final velocity, what will be the new position of the bottom of the block at time t = 2 × 0.09 = 0.18seconds?
y = ? m

I understand everything up until this second step position update. I am not sure how to compute it. My first try I got 0.478478 m but that was incorrect. I appreciated if I could have some guidance.

Welcome to physicsforums!

Troi Jones said:
Fspring= kx
Fgrav= ma
Fnet= Fspring + Fgrav
p= mv
delta p= F*delta t
Good job at the equations.
Troi Jones said:
update position= Vavg*delta t= delta r
Troi Jones said:
Momentum update: Just after releasing the block, the momentum of the block is zero. Approximate the average net force during the next time interval by the force you just calculated. At t = 0.09 seconds, what will the new momentum and velocity of the block be?
py = 0.145197 kg · m/s
vy = 2.074 m/s

Position update: Initially the bottom of the block is at y = 0.12 m. Approximating the average velocity in the first time interval by the final velocity, what will be the new position of the bottom of the block at time t = 0.09 seconds?
y = 0.30666 m
Here's the thing, you found that the velocity at t=0.09 seconds was vy=2.074 m/s. When you started to calculate the position update, you assumed the same final velocity(2.074 m/s). But this is the velocity at t=0.09, not t=0; at t=0 the particle started with 0 velocity.

So comes the remedy: stop using velocity to calculate position update. Use the force equations instead.

phoenix95 said:
Welcome to physicsforums!Good job at the equations.Here's the thing, you found that the velocity at t=0.09 seconds was vy=2.074 m/s. When you started to calculate the position update, you assumed the same final velocity(2.074 m/s). But this is the velocity at t=0.09, not t=0; at t=0 the particle started with 0 velocity.

So comes the remedy: stop using velocity to calculate position update. Use the force equations instead.
Sorry, I am seem to be a bit confuse with the statement of using the force equations to find the new position than using velocity. Do I have to manipulate the force equations and set them equal to delta r?

Troi Jones said:
Do I have to manipulate the force equations and set them equal to delta r?
Force equations are fine, no need to manipulate them again.

Troi Jones said:
Fspring= kx
(I didn't notice before but the sign is wrong)
Use this equation. K is a constant forever so only variables are Fspring and x. But this time note that the force that is causing the spring length to change is actually the weight of the mass that is sitting on top of the spring.

Write ΔF = -kΔx, and continue.

phoenix95 said:
Force equations are fine, no need to manipulate them again.

(I didn't notice before but the sign is wrong)
Use this equation. K is a constant forever so only variables are Fspring and x. But this time note that the force that is causing the spring length to change is actually the weight of the mass that is sitting on top of the spring.

Write ΔF = -kΔx, and continue.
Ohhhhhh, that makes much more sense now. Thank you very much!

phoenix95 said:
Force equations are fine, no need to manipulate them again.

(I didn't notice before but the sign is wrong)
Use this equation. K is a constant forever so only variables are Fspring and x. But this time note that the force that is causing the spring length to change is actually the weight of the mass that is sitting on top of the spring.

Write ΔF = -kΔx, and continue.
Is this how my equation is suppose to be: ΔF = -kΔx = (0.43-2.3)= -(10)(xfinal - 3.0666)?

## 1. How do I find the position update for Step 2 in the Spring Problem?

To find the position update for Step 2 in the Spring Problem, you will need to use the equation x(t) = x0 + v0t + ½at2, where x0 represents the initial position, v0 represents the initial velocity, and a represents the acceleration. Plug in the values for these variables and the time t to solve for the position update.

## 2. What is the significance of Step 2 in the Spring Problem?

Step 2 in the Spring Problem is important because it represents the position update of an object at a specific time t. This step helps us track the movement and position of the object over time, which is crucial in understanding the behavior of the spring system.

## 3. How does the position update in Step 2 affect the overall problem?

The position update in Step 2 directly affects the overall problem by showing us the change in position of the object at a specific time t. This information is necessary for solving the entire problem and understanding the dynamics of the spring system.

## 4. What factors can influence the position update in Step 2 of the Spring Problem?

The position update in Step 2 can be influenced by various factors such as the initial position and velocity of the object, the acceleration due to the spring force, and any external forces acting on the object. Other factors such as air resistance and friction may also play a role in the position update.

## 5. Is there a specific formula to find the position update in Step 2 of the Spring Problem?

Yes, there is a specific formula to find the position update in Step 2 of the Spring Problem. As mentioned earlier, it is x(t) = x0 + v0t + ½at2. However, the values for the variables may vary depending on the specific problem and its given parameters.

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