How far does a block travel before encountering a spring?

[preluderacer]

Homework Statement

A block starts from rest at the top of a 31.0° inclined plane and encounters a spring, of constant 3.4 E3 N/m rigidly attached to the plane. If the block's mass is 33.0 kg and it compresses the spring by 37.0 cm, find the distance the block traveled before it encountered the spring.

Homework Equations

0.5(kx^2) i think

The Attempt at a Solution

Im quite sure how to find the speed or distance of the block.

 

[tiny-tim]

Hi preluderacer!

Im quite sure how to find the speed or distance of the block.

Excellent!

So you're using conservation of energy (KE + PE = constant)? ​

 

[preluderacer]

Ok here's what I did. Tell me if this was the correct thing to do. I set (1/2)mv^2=(1/2)kx^2 then solved for v. I got 3.76 m/s. After that I found out the acceleration by the block by doing 9.8cos31 degrees. I got 5.04 m/s^2. I then found out by doing 3.76/5.04 = 0.746 seconds. I then used the equation 1/2(V-Vo)(t). I found the distance to be 1.40 meters. Does this sound right?

 

[tiny-tim]

hi preluderacer!

(try using the X² icon just above the Reply box )

Ok here's what I did. Tell me if this was the correct thing to do. I set (1/2)mv^2=(1/2)kx^2 then solved for v. I got 3.76 m/s. After that I found out the acceleration by the block by doing 9.8cos31 degrees. I got 5.04 m/s^2. I then found out by doing 3.76/5.04 = 0.746 seconds. I then used the equation 1/2(V-Vo)(t). I found the distance to be 1.40 meters. Does this sound right?

Not quite.

You haven't included the gravtiational PE lost between the block hitting the spring (at speed v) and coming to rest.

And is it cos?

 

[preluderacer]

I did do 9.8sin30 degrees I just wrote cos for some reason here hah. Do I subject mgh then? I am confused now ughh.

 

[tiny-tim]

Do I subject mgh then? I am confused now ughh.

Do you subject?

Do you object?

Your 1/2 mv² = 1/2 kx² needs an extra mg something!

 

[preluderacer]

Do I subtract mgh from the 1/2kx^2?

 

[tiny-tim]

work it out …

if there's more x, ie further downhill, does the gravity (forgetting the spring) tend to make v less or more?

 

[preluderacer]

I thought I got V by doing 9.8sin31?

 

[tiny-tim]

only until it hits the spring …

once it starts compressing the spring, v will also be affected by the spring

look at it another way … once you get the speed when it hits the spring, obviously from then on the spring tends to make the speed less, but does gravity tend to make the speed less or more?

 

[preluderacer]

I does more. So do I need to add mg?

 

[tiny-tim]

you need to write the equation so that increasing x increases v …

so if they're on opposite sides of the =, they must have the same sign

 

[preluderacer]

mgh=1/2kx^2+1/2mv^2? I am so lost.

 

[tiny-tim]

mgh=1/2kx^2+1/2mv^2?

yes, and now write h is terms of x

 

[preluderacer]

So do I just divide by mg?

 

[tiny-tim]

hi preluderacer!

(just got up :zzz: …)

So do I just divide by mg?

no, rewrite h in terms of d (the distance asked for in the question), x, and 31°…

that will give you an equation for d, which is what you need!