PE and KE, mass sliding down incline into a spring (no friction)

[J-dizzal]

Homework Statement

In the figure, a block of mass m = 15 kg is released from rest on a frictionless incline of angle θ = 26°. Below the block is a spring that can be compressed 3.1 cm by a force of 380 N. The block momentarily stops when it compresses the spring by 4.0 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/fig08_41.gif

Homework Equations

The Attempt at a Solution

 

[Nathanael]

Look at your expression for W_s at the bottom... you used (0.4m)^2 instead of (0.04m)^2

 

[J-dizzal]

Hey Nathanael, thanks. let me try this again

 

[J-dizzal]

Look at your expression for W_s at the bottom... you used (0.4m)^2 instead of (0.04m)^2

when solving for part b I am using K=1/2 mv² , where K=9.8065, m=15 and i get v=1.143

 

[J-dizzal]

using kinematic eq. v²=2a(x-x₀) where a=4.296 and x=0.15218. I got v=0.4461 m/s but not correct

 

[Nathanael]

using kinematic eq. v²=2a(x-x₀) where a=4.296 and x=0.15218. I got v=0.4461 m/s but not correct

Kinematics equations were derived with the assumption that acceleration is constant, so it doesn't apply to this situation. (Their use is pretty limited.)

when solving for part b I am using K=1/2 mv² , where K=9.8065, m=15 and i get v=1.143

I think this is the trap they expected you to fall into. Be careful with the energy: is there any other form of energy involved?

 

[J-dizzal]

Kinematics equations were derived with the assumption that acceleration is constant, so it doesn't apply to this situation. (Their use is pretty limited.)I think this is the trap they expected you to fall into. Be careful with the energy: is there any other form of energy involved?

I don't see why kinematic eq won't work, because its asking for velocity before it even touches the spring. so a should be constant?
There could be a potential energy of gravity on the block. U=mgh

 

[Nathanael]

I don't see why kinematic eq won't work, because its asking for velocity before it even touches the spring. so a should be constant?

You're right, I wasn't thinking about what you were doing sorry
Your mistake is that x you used is the total distance including the distance compressed. You just want the distance before it touches.

There could be a potential energy of gravity on the block. U=mgh

Right. As the spring is compressed gravity is still doing work, so the kinetic energy of the block right before it touches is not equal to the total energy of the system.

 

[J-dizzal]

You're right, I wasn't thinking about what you were doing sorry
Your mistake is that x you used is the total distance including the distance compressed. You just want the distance before it touches.Right. As the spring is compressed gravity is still doing work, so the kinetic energy of the block right before it touches is not equal to the total energy of the system.

wouldnt the total distance be .15218+.04?

Edit. nevermind its .15218-.04 for the dist before spring

Edit2. after plugging in .11218 i still am doing something wrong when solving for v

Edit3. ok i got it. i was trying to square .11218 in the kinematic equation, i don't know why i did that.

 

[Nathanael]

Since you got it I will tell you the energy method of doing it: $E_{kinetic}=E_{total}-0.04mg\sin(26°)$ and you already determined E_total for an earlier part.

What you used in post #4 was E_kinetic=E_total