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

[Swan]

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

Homework Statement

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

Homework Equations

F_N = m*g*cosθ
F_f = μ*F_N
W = F*Δd
E_T1 - W_f = E_T2
Let H represent the height off the ground. Let h represent height between the boxes.

The Attempt at a Solution

V₁ = 0. So initial kinetic energy is negligible.
h = (sin67°)(1.3m + 0.4m)
W_f = F_f * (1.3m + 0.4m)
W_f = 6.9J

E_T1 - W_f = E_T2
mgH + $\frac{1}{2}$k*x² - W_f = mg(h+H) + $\frac{1}{2}$mV²₂
mgH + $\frac{1}{2}$k*x² - W_f = 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?

 

[Doc Al]

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

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.)

 

[Swan]

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.)

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.

E_T1 - W_f = E_T2
mgH + $\frac{1}{2}$k*x² - W_f = mg(h+H) + $\frac{1}{2}$mV²₂
mgH + $\frac{1}{2}$k*x² - W_f = 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.

 

[Doc Al]

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.

I understand.

When I carry out the solution, it won't work because of the negative and square root result in error.

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.)

 

[Swan]

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.)

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?

 

[Doc Al]

I tried it. It still yields a negative which will result in a error through square root.

Yep.

Is there any error in the derivation of my equation?

No. Looks like the problem is bogus.

What book is it from?

 

[Swan]

Yep.

No. Looks like the problem is bogus.

What book is it from?

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

 

[Doc Al]

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

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

 

[Swan]

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

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.

 

[Doc Al]

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.

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

 

[Swan]

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

Ok, thank you.

 

[Doc Al]

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.

 

[Swan]

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.

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

 

[Doc Al]

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

Sure.