Calculating Final Velocity with Spring and Friction on Inclined Plane

[bmoore509]

Homework Statement

A spring with a spring-constant 2.9 N/cm is compressed 34 cm and released. The 9 kg mass skids down the frictional incline of height 30 cm and inclined at a 24◦ angle. The acceleration of gravity is 9.8 m/s2 . The path is frictionless except for a distance of 0.8 m along the incline which has a coefficient of friction of 0.2.
What is the final velocity vf of the mass?
Answer in units of m/s.

k=2.9
d=34 cm
m= 9
h=30cm
theta=24
distance of friction = .8 meters
Coefficient of friction=.2

Homework Equations

N=mgcostheta
Ff=Nk
Us+Ug+Wnc=K

The Attempt at a Solution

So I was looking up similar problems and this is what someone was told to do and I tried it but it didn't work at all. Can you tell me where I went wrong?

N=(9)(9.8)(cos24)=82.57
Ff=Nk=233.67

Us+Ug+wnc=K
.5kx^2+mgh+Ff(df)=.5mv^2
v=31.65

Which wasn't right.

 

[Filip Larsen]

The first two numbers you quote in 3) is not correct. In the first number you have probably just made a typo, but the second (233.67) seems way off. If you multiply around 80 with 0.2 you do not get 233.

Also, not all terms in your energy conversation equation have correct sign.

 

[fawk3s]

Which wasn't right.

This is leading me to think you already know the right answer. Could it be about 2,6 m/s? Because if it is not, then there's no point in me hinting you with incorrect values.

 

[bmoore509]

It's online homework which means I have so many tries to get it right. I think I've already used 3/7 of the tries.

I totally used the wrong K. Oh my gosh. Ugh.
N=(9)(9.8)(cos24)=80.57
Ff=Nk=16.11494187

But there's no point in me continuing from there if my next equation is wrong. How do I know which signs are wrong?

 

[Filip Larsen]

You seem to have found all the correct energy terms, so you just need to think about which energy terms concern the situation before the spring is released and equate the sum of those to the sum of terms that concern the situation just after the mass has reached the end of the incline. Hint: the two situations each has two terms that apply to it.

 

[fawk3s]

.5kx^2+mgh+Ff(df)=.5mv^2

Looking at this equation, the only thing I can think of you can do wrong, besides miscalculating, is that you actually ADD the job done by the friction. If that's so, think about the direction of the force applied by the friction.

 

[bmoore509]

So it would actually be
Us+Ug+wnc=K

Well, if change in K + change in U = Wnc. Then
.5mvf^2-.5mvi^2+.5kxf^2-.5mxi^2=Ff(df)

vi=0 so we get
.5mvf^2+.5kxf^2-.5mxi^2=Ff(df)

But then does that mean the second U should actually be the mgh? In which case, I'd get

.5(9)v^2+.5(2.9)(34^2)-(9*9.8*34)=16.11494187*0.8
4.5v^2+(1676.2)-(2998.8)=12.8919535
4.5v^2=1335.491954
v=17.22718751 m/s


If I messed up, can you tell me where?

 

[Filip Larsen]

You seem to mess up early on, seemingly by inserting the numbers relevant for your problem into an equation that (maybe?) applies to another type of problem.

I recommend that you try understand the flow of energy relevant for your problem by thinking about what contributes to the total energy at different times. Try write up, one term at a time, the energies involved initially before the mass is release (hint: two terms is needed, one for the potential energy stored in the spring and one for the potential energy from gravity). Then write up the final energy terms after sliding (hint: two terms needed, one for the kinetic energy of the mass and one for the frictional loss). Now sum the initial energy terms and equate them to the sum of the final energy terms.

If you have trouble seeing why a particular term is needed or not, you can also simply write up all relevant types of energy terms (spring potential, gravitational potential, mass kinetic, frictional loss) for both initial and final situation, equate their sums (that is, apply energy conservation), and then insert values for each of the 8 terms (of which 4 are zero) and solve for final velocity.

 

[bmoore509]

So with that information, the part I got wrong then is that it should be -Ug.

So Us-Ug=K-Wnc.

Correct?

 

[Filip Larsen]

You got the correct type of terms on left and right side of the equation, but for reasons that escape me you have selected minus sign for two of them. All four terms of energy in your last equation should be positive.

 

[bmoore509]

So it's not minus even though that's gravity and friction? The only issue in my original part was that since the Wnc had been moved over (as a positive on the right), it needed to be negative?

 

[bmoore509]

So what I should get is .5kx^2+mgh=.5mv^2+Ff*df
(I realize in previous problems I kept forgetting to convert 34 cm and 30 cm to meters.)

.5(2.9)(.034^2)+9(9.8)(.030)=.5(9)v^2+16.11494187(.8)
0.0016762+2.646=4.5v^2+12.8919535
-10.2442773=4.5v^2

which isn't right because you can't square root a negative. So do you not convert?

 

[bmoore509]

Oops. My math skills went dumb.

.5(2.9)(.34^2)+9(9.8)(.3)=.5(9)v^2+16.11494187(.8)

0.16762+26.46=4.5v^2+12.8919535
13.7356665=4.5v^2
v^2=3.052370333
v=1.747103412

Which isn't right.

 

[fawk3s]

Ofcourse it isn't right. You got the units randomly mixed together. You have converted the length of the spring nicely into meters, yet your springs elasticity constant is still N/cm.

 

[bmoore509]

5(290)(.34^2)+9(9.8)(.3)=.5(9)v^2+16.11494187(.8)

167.62+26.46=4.5v^2+12.8919535
194.08=4.5v^2+12.8919535
v=6.34539

Which still isn't right. I converted 2.9 to 290 and my answer is still wrong. I only have two tries left. What am I doing wrong?

 

[fawk3s]

Calculation error. Check your numbers closely.

 

[bmoore509]

Thank you. I need to be more careful with my work. Ugh.


But thank you very much.