Length along ramp from end of spring, picture included

[gap0063]

Homework Statement

A(n) 2300 g block is pushed by an external
force against a spring (with a 14 N/cm spring
constant) until the spring is compressed by
19 cm from its uncompressed length. The
compressed spring and block rests at the bot-
tom of an incline of 33◦ .
The acceleration of gravity is 9.8 m/s2 .
Note: The spring lies along the surface of
the ramp (see figure).
Assume: The ramp is frictionless.
Now, the external force is rapidly removed
so that the compressed spring can push up
the mass.
[PLAIN]http://img64.imageshack.us/img64/5812/problem6l.jpg
(a)How far up the ramp (i.e., ℓ the length
along the ramp) will the block move (measured
from the end of the spring when the
spring is uncompressed) before reversing di-
rection and sliding back?
Remember, the block is not attached to the
spring.
Answer in units of cm.

(b)Repeat the first part of this question, but
now assume that the ramp has a coefficient of
friction of μ = 0.308 .
Keep all other assumptions the same.
Answer in units of cm.

Homework Equations

[tex]\Delta[/tex]K= K_f-K_i=1/2mv_f²-1/2mv_i²
E_total=1/2mk²+1/2mv²

The Attempt at a Solution

I don't know where the angle comes in...
I know that N=mgcos[tex]\theta[/tex]

 

[PhanthomJay]

Homework Statement

A(n) 2300 g block is pushed by an external
force against a spring (with a 14 N/cm spring
constant) until the spring is compressed by
19 cm from its uncompressed length. The
compressed spring and block rests at the bot-
tom of an incline of 33◦ .
The acceleration of gravity is 9.8 m/s2 .
Note: The spring lies along the surface of
the ramp (see figure).
Assume: The ramp is frictionless.
Now, the external force is rapidly removed
so that the compressed spring can push up
the mass.

(a)How far up the ramp (i.e., ℓ the length
along the ramp) will the block move (measured
from the end of the spring when the
spring is uncompressed) before reversing di-
rection and sliding back?
Remember, the block is not attached to the
spring.
Answer in units of cm.

(b)Repeat the first part of this question, but
now assume that the ramp has a coefficient of
friction of μ = 0.308 .
Keep all other assumptions the same.
Answer in units of cm.

Homework Equations

[tex]\Delta[/tex]K= K_f-K_i=1/2mv_f²-1/2mv_i²
E_total=1/2mk²+1/2mv²

The Attempt at a Solution

I don't know where the angle comes in...
I know that N=mgcos[tex]\theta[/tex]

Note that the block starts off with no speed, and it has no speed when it comes to a stop at the top of the incline, so delta K =0 . Note also that your energy equation is wrong. W_nc = (delta K + delta U_spring + delta U_gravity), where W_nc = work done by non conservative forces, like friction.

 

[gap0063]

Note that the block starts off with no speed, and it has no speed when it comes to a stop at the top of the incline, so delta K =0 . Note also that your energy equation is wrong. W_nc = (delta K + delta U_spring + delta U_gravity), where W_nc = work done by non conservative forces, like friction.

So

W_nc= 1/2kx²+mgh

is k the 19cm (.19m)?

and do I figure out the height from the angle? and if so how?

 

[AHinkle]

okay k is the spring constant which is given above 14 N/cm or 1400N/m if you convert it like I did.

so you know that initial potential energy is the potential energy in the spring plus the gravitational potential energy. If you set you reference point there you can say that gravitational potential energy there is 0 because in mgh, h=0.

and we also know that when we get to the top of the slope K=0 and U=mgh
so use the information up to there to solve for the height.

if you've got the max height you can set up a triangle (sin = opp/hyp, and you know the angle and the opposite side) and solve for the distance up the ramp, but the distance you get from that is going to be the distance from the top of the "compressed" spring so compensate for that in your answer. (i.e. subtract the distance that the spring was compressed)

 

[gap0063]

okay k is the spring constant which is given above 14 N/cm or 1400N/m if you convert it like I did.

so you know that initial potential energy is the potential energy in the spring plus the gravitational potential energy. If you set you reference point there you can say that gravitational potential energy there is 0 because in mgh, h=0.

and we also know that when we get to the top of the slope K=0 and U=mgh
so use the information up to there to solve for the height.

if you've got the max height you can set up a triangle (sin = opp/hyp, and you know the angle and the opposite side) and solve for the distance up the ramp, but the distance you get from that is going to be the distance from the top of the "compressed" spring so compensate for that in your answer. (i.e. subtract the distance that the spring was compressed)

would it be
1/2kx²=mgh
h=kx²/2mg

well i tried and did 1400(.19)²/(2*2.3*9.8)
h=1.12112m

then sin33=o/h
h=o/sin30
h=2.05846

2.05846-.19 m=1.8684m

which is 186.84 cm which is wrong

well I reread the problem and it says when it is uncompressed so .57 m
I misused 2.05846-.57=1.48846

which is 148.46 cm which is wrong

 

[AHinkle]

We have the same homework service I am pretty sure... mine was due today so here's how I did it.

Since there is no friction on the first part of the problem, mechanical energy is conserved and you can use U_i+K_i=U_f+K_f

U_i= (1/2kx² + mgh_i) where h_i = 0
I put the reference point a the top of the compressed spring.
x is 19cm or 0.19m the distance the spring was compressed

K_i = 1/2mv_i²
but the block is at rest so v=0 so the whole quantity is 0
so K_i=0

U_f = (mgh)
(only mgh because there's no spring potential energy like there was before)

K_f = 1/2 mv_f²
but again at the top of the motion the velocity is 0 so
K_f = 0

so...
U_i+K_i=U_f+K_f
(1/2kx²+0) + 0 = mgh_f + 0

1/2kx² = mgh_f
solve for h_f

(1/2kx²)/(mg)=h_f
(kx²/2) x (1/mg) = h_f

h_f=(kx²/2mg)
convert everything to meters btw, I don't know why I do it .. i think it has to do with g being
in units of m/s², and i converted the mass to kilos

k=1400 N/m (14 N/cm)
x=.19m (19cm)
m=2.3Kg (2300g)

h_f=((1400N/m)(.19m)²)/(2(2.3Kg)(9.8m/s²))

h_f=1.1211 Meters (112.1118 cm)

[tex]\theta[/tex] = 33 degrees
sin(33) = (h_f)/d
let d be the distance from the top of the compressed spring to the top of the motion

(d)sin(33) = h_f
d = h_f/sin(33)

d=112.1118/sin(33) = 205.8460cm

L = d-x (this is the x from earlier, or the distance the spring was compressed)

L = 205.8460 cm - 19cm = 186.846 cm

L = 186.8460 cm

the part with friction is similar but you have to account for energy lost to internal energy

 

[AHinkle]

then sin33=o/h
h=o/sin30
h=2.05846

did you actually do sin(30) ? maybe you just messed up the angle it looks like you did it right for the most part..