# * * Elastic potential energy with box going up an incline.

1. Apr 22, 2012

### Swan

*URGENT* Elastic potential energy with box going up an incline.

1. The problem statement, all variables and given/known data
A spring having a force constant of 240 $\frac{N}{m}$ is placed on a plane inclined at 67° to the horizontal. The spring is compressed 0.40 m and a 2.0 kg mass is placed on it. The coefficient of kinetic friction along the inclined surface is 0.53. Find the speed of the mass after it moves 1.3 m up the plane (above the completely extended spring).

Diagram

2. Relevant equations
FN = m*g*cosθ
Ff = μ*FN
W = F*Δd
ET1 - Wf = ET2
Let H represent the height off the ground. Let h represent height between the boxes.

3. The attempt at a solution
V1 = 0. So initial kinetic energy is negligible.
h = (sin67°)(1.3m + 0.4m)
Wf = Ff * (1.3m + 0.4m)
Wf = 6.9J

ET1 - Wf = ET2
mgH + $\frac{1}{2}$k*x2 - Wf = mg(h+H) + $\frac{1}{2}$mV22
mgH + $\frac{1}{2}$k*x2 - Wf = mgH + mgh + $\frac{1}{2}$mV22
**Both mgH cancel out on each sides**
$\frac{1}{2}$(240$\frac{N}{m}$)(0.4m)2 - 6.9J = [(2kg)(9.8m/s2)(sin67°)(1.3m + 0.4m)]+ $\frac{1}{2}$(2kg)V22

I worked it out but I'm receiving an error. What am I doing wrong?

2. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Looks OK to me. But that diagram is a bit ambiguous about the meaning of the 1.3 m distance. (It doesn't reference the compression and seems to imply that the total distance moved is 1.3 m.)

3. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

Firstly, thank you so much for replying quickly :).

In the question it says "(above the completely extended spring)" meaning the the box has to move 1.3m after full extension. At the moment the box is on the spring, it is compressed 0.4m and therefore total distance becomes (1.3m + 0.4m = 1.7m). This was the way I interpreted it. When I carry out the solution, it won't work because of the negative and square root result in error.

Now to work out the whole equation.

ET1 - Wf = ET2
mgH + $\frac{1}{2}$k*x2 - Wf = mg(h+H) + $\frac{1}{2}$mV22
mgH + $\frac{1}{2}$k*x2 - Wf = mgH + mgh + $\frac{1}{2}$mV22
**Both mgH cancel out on each sides**
$\frac{1}{2}$(240$\frac{N}{m}$)(0.4m)2 - 6.9J = [(2kg)(9.8m/s2)(sin67°)(1.3m + 0.4m)]+ $\frac{1}{2}$(2kg)V22
12.3J = 30.67J + $\frac{1}{2}$(2kg)V2
-18.39J = $\frac{1}{2}$(2kg)V2

Resulting in error.

4. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

I understand.
Yes. Obviously the data is bogus, given that interpretation of the problem. Try solving it where 1.3 m is the total distance traveled up the ramp. (See if that at least presents possible data.)

5. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

I tried it. It still yields a negative which will result in a error through square root. Is there any error in the derivation of my equation?

6. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Yep.
No. Looks like the problem is bogus.

What book is it from?

7. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

Its from the Nelson Physics 12 but I think the question was modified by the teacher.

8. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Hmm... how is the problem stated in the book? Perhaps your teacher made an error when modifying it.

9. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

The original question with different values was going down the incline in which all worked out. But then when teacher modified when making it go up the incline, I'm getting an error.

10. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Yes. The modified problem statement is bogus. The data represents a physically impossible situation.

11. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

Ok, thank you.

12. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Just for fun, why not try calculating the maximum distance up the incline the mass will reach? (Using the given compression, spring constant, and μ.) See if that at least gives a sensible answer.

13. Apr 22, 2012

### Swan

Re: *URGENT* Elastic potential energy with box going up an incline.

I would let the system have 0 kinetic energy at the top to reach its max height.

14. Apr 22, 2012

### Staff: Mentor

Re: *URGENT* Elastic potential energy with box going up an incline.

Sure.