Why does Hooke's law not work here?

AI Thread Summary
The discussion centers on the application of Hooke's law in a scenario involving a car and a coiled spring. The initial method attempted to calculate the spring constant using average acceleration and force, leading to an incorrect value due to the assumption of constant acceleration. It was clarified that Hooke's law applies to the force exerted by the spring when stretched, not the average force during deceleration. The correct approach involves using the conservation of energy principle, equating kinetic energy to potential energy stored in the spring. The misunderstanding arose from misapplying the kinematic equation, which is not valid when forces are variable.
MightyDogg
Messages
14
Reaction score
0

Homework Statement


  1. A 1200-kg car moving on a horizontal surface has speed v = 85 kmh when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Homework Equations


F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2

The Attempt at a Solution


I tried to find the average acceleration to slow the car from 85kmh to 0. I used the formula vfinal^2=vinitial^2 + 2ax, where velocity initial is 23.6m/s and x is 2.2m. This gave me an acceleration of -126m/s^2. Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

However, I am supposed to use the conservation of energy principle where KE=PE. This gives the correct answer. Why does my method not work?
 
Physics news on Phys.org
MightyDogg said:

Homework Statement


  1. A 1200-kg car moving on a horizontal surface has speed v = 85 kmh when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Homework Equations


F=-kx
KE=(1/2)mv^2
PE(spring)=(1/2)kx^2

The Attempt at a Solution


I tried to find the average acceleration to slow the car from 85kmh to 0. I used the formula vfinal^2=vinitial^2 + 2ax, where velocity initial is 23.6m/s and x is 2.2m. This gave me an acceleration of -126m/s^2. Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

However, I am supposed to use the conservation of energy principle where KE=PE. This gives the correct answer. Why does my method not work?

I does not work because the acceleration is not constant. Hook's Law works, but not your expression ##v_f^2 = v_i^2 + 2ax##.
 
Last edited:
MightyDogg said:
Then I multiplied the acceleration by the mass to find the average force. This gave me -151200N. Then, I plugged that into Hooke's law with x being 2.2m. This gave the spring constant being 69000N/m.

By Hooke's law I assume you mean ##F=kx## where ##F## is the magnitude of the force and ##x## is the distance?

In that formula ##F## is not the average force. It's the magnitude of the force when the spring is stretched (or compressed) a distance ##x##.

If the force were constant, that would work, but the force is not constant. You could integrate the force, or use energy concepts.
 
Oh, okay it makes sense now. Thank you both very much.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top