Relating to potential energy of a spring and mass on inclined plane

AI Thread Summary
The discussion centers on calculating the compression of a spring when a block on an inclined plane comes to rest. Given parameters include a spring constant of 525 N/m, a block mass of 2.71 kg, and an incline angle of 20 degrees. Participants emphasize using conservation of energy, equating gravitational potential energy and elastic potential energy, while clarifying the role of kinetic energy at different points. The correct approach involves considering the vertical displacement and the initial kinetic energy of the block, which is not zero. The final answer for the spring compression is stated as x = 0.143 m, indicating the need for accurate calculations in energy conservation.
NewSoul
Messages
14
Reaction score
0

Homework Statement


Here is an illustration to ease understanding:
7-p-063.gif

Angle inclined plane: θ = 20.0°
spring of force constant k = 525 N/m
block: m = 2.71 kg
distance of block from spring: d = 0.330 m
speed of block: v = 0.750 m/s

By what distance is the spring compressed when the block momentarily comes to rest? (I'm going to call it 'x' in this problem.)

The answer should be x = 0.143 m

Homework Equations


Ok, I'm not quite sure how many of these are relevant or if I'm missing some, but I'll just give some that I know.

PEs = (1/2)kx2 (x = distance spring is compressed)
PEg = mgh
KE = (1/2)mv2
F = ma
U + KE = 0 (U = any potential energy? All potential energies? I'm not quite positive what is going on here.)

The Attempt at a Solution


Okay. I'm not quite sure how to go about this, so I'll just start doing what I seem to know.

Let's start with force. It doesn't seem inherently relevant to the problem, but why not?

Okay, so force in the x direction seems to be: mgsinθ = ma. So we get a = gsinθ = (9.8)(sin 20°) = 3.35198 m/s2.

Force in the y direction seems to be: N - mgsinθ = 0. So N = mgsinθ = (2.71)(9.8)(sin 20°) = 9.08337 N.

-------------

Ok. That was probably not relevant. Let's do work now.

Using the numbers above, I get KE = (1/2)(2.71)(.750)2 = 0.76219 J.

PEg = (2.71)(9.8)h. Okay, I'm not quite sure if h should just be d or if I have to factor in the slope. I'm thinking that slope probably does have to be included, so h is 0.330(sin 20°), so PEg is 2.99751 J.

So it seems like PEg + PEs + KE = 0. Then it must be that (1/2)kx2 + mgh + (1/2)mv2 = 0... However, solving for x gives me the square root of a negative number, I must be interpreting U + KE = 0 wrong. Where do I go from here? Thanks, I know this is long.
 
Physics news on Phys.org
Why do you think that the energy is zero?
You can use conservation of energy to solve the problem.
Equate the total energy in the initial position with the total energy in the final position.
Total energy here includes kinetic, potential gravitational and potential elastic.
Some terms may be zero, like elastic energy in the initial position or kinetic energy in the final position.
 
nasu said:
Why do you think that the energy is zero?
Well, I'm not entirely sure what is meant by the equation U + KE = 0 or if it's relevant...But I do know that a decrease in potential energy would mean an increase in kinetic energy and vice versa.

You can use conservation of energy to solve the problem.
Equate the total energy in the initial position with the total energy in the final position.
Total energy here includes kinetic, potential gravitational and potential elastic.
Some terms may be zero, like elastic energy in the initial position or kinetic energy in the final position.

So are you saying that PEg = PEs (KE is zero at the initial and final positions)? I'm afraid I don't understand.

*edit* Hold on a minute I did something weird...

Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?
 
Last edited:
NewSoul said:
Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?
h is the total vertical movement of the block. Don't forget that this will include a contribution from x.
 
NewSoul said:
So are you saying that PEg = PEs (KE is zero at the initial and final positions)? I'm afraid I don't understand.

*edit* Hold on a minute I did something weird...

Okay, so if PEg = PEs, then I get (1/2)kx2 = mgh... and x = √(2mgh/k) = 0.107 m, which is incorrect. If I assume h shouldn't have sin in it, then I get x = 0.183 m. Both of these answers are still wrong. Shouldn't I have to fit velocity into the problem somehow?

No, the initial KE is not zero. It moves with some speed, isn't it?
And h is measured along the vertical direction.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top