Loop-the-loop on frictionless track, spring-powered

[fysics4liefe]

Homework Statement

A toy car of mass 0.300 kg is shot out of a compressed horizontal spring (A), and then travels around a loop-the-loop along a frictionless track. The spring has a spring constant k = 900 N/m, and 3.50 J of energy was stored in the spring by compressing it.

(1) How far was the spring compressed to store 3.5 J of energy in it?
(2) What is the speed of the car as it passes point B, which is 0.400 m above point A?
(3) How high above the original position does the car go before stopping (point C)?
(4) Suppose now that the track has some friction, and that it is observed that the car's height at point C is only 80% of the height it would reach in the absence of friction (i.e. in Question #3). How much energy did friction remove on the trip from A to C?

Homework Equations

KE = (1/2)mv²
PE = mgh
PE_S = (1/2)kx²

The Attempt at a Solution

(1) PE = (1/2)kx², so 3.5 = (0.5)(900)x². Solving for x, x = 0.088. So this is the distance the spring was compressed in order to store 3.5 J.
(2) To be honest, I have attempted this about 10 different ways, none of which were correct. I believe that mgh = (1/2)mv² could be used in some way, but am not sure of how to use this, because I don't know what the maximum height would be. (1/2)mv₁² = (1/2)mv₂² + mgh could be useful as well, if either velocity was known. Perhaps the velocity can be figured from the amount of energy that was stored in the spring.
(3) & (4) Without knowing the answer to #2, it is difficult/impossible to even get started on these ones. Please advise. Thank you all in advance.

 

[fysics4liefe]

(2) Another attempt... PE(spring) = PE = (1/2)mx² ==> (1/2)(900)(0.088^2) = 3.4848. Plugging this value of PE into PE = (1/2)mv², I get V(initial) = 4.82. Plugging this into (1/2)mV^2 = (1/2)mV^2 + mgh, I get that the velocity at the top of the ramp = 3.92 m/s. Is this sensible?

 

[PhanthomJay]

(2) Another attempt... PE(spring) = PE = (1/2)[strike]m[/strike]kx² ==> (1/2)(900)(0.088^2) = 3.4848. Plugging this value of PE into PE = (1/2)mv², I get V(initial) = 4.82.

yes, this is good

Plugging this into (1/2)mVinitial^2 = (1/2)mVfinal^2 + mgh, I get that the velocity at the top of the ramp = 3.92 m/s. Is this sensible?

using h = 0.4 m, you get the velocity at that point B. To get the max height when the car stops, what is v_final and what is h at that point C?

 

[fysics4liefe]

yes, this is good using h = 0.4 m, you get the velocity at that point B. To get the max height when the car stops, what is v_final and what is h at that point C?

Okay, got it now, thank you. So the energy at the highest point is the sum of KE + PE, which is (1/2)mv² + mgh. Since v = 0 at the highest point, (1/2)mv^2 = 0. So kinetic energy is zero (obviously, since we aren't moving). So the total energy (which is conserved on frictionless track) is equal to the PE, mgh. (0.3 kg)(9.8 m/s^2)(h) = 3.4848. Solving for h at v=0, we get h = 1.185 meters, which is the max height (correct significant figures notwithstanding).

Which enables us to solve question 4. 80% of previous max. height is 0.952 meters, since (0.8)(1.185) = 0.952. Plugging this into mgh, we get a total PE of 2.7988 J. On frictionless track, this number was 3.4848. So, 3.488 J (frictionless energy) - 2.7988 J (energy after friction) = 0.69 J.

Does that all look sound? Thank you, PhanthomJay. It is much appreciated.

 

[PhanthomJay]

Yes, looks very good!