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

[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.

 

[phoenix95]

Welcome to physicsforums!

Fspring= kx
Fgrav= ma
Fnet= Fspring + Fgrav
p= mv
delta p= F*delta t

Good job at the equations.

update position= Vavg*delta t= delta r

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 v_y=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.

 

[Troi Jones]

Welcome to physicsforums!Good job at the equations.Here's the thing, you found that the velocity at t=0.09 seconds was v_y=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?

 

[phoenix95]

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.

Fspring= kx

(I didn't notice before but the sign is wrong)
Use this equation. K is a constant forever so only variables are F_spring 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.

 

[Troi Jones]

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 F_spring 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!

 

[Troi Jones]

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 F_spring 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)?