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VERY basic work-energy question

Started by Nervo Mar 2, 2004 at 2:39 PM 3 replies 700+ views

Original Post

Mar 2, 2004 at 2:39 PM

I had a physics question that I was able to solve correctly using the work-energy theorem, and thats fine and dandy, but conceptually I can't help but thinking that I'm missing something. the question was:

quote:
A 1000 kg car being tested for crashworthiness is driven at 20 m/s into a giant spring of force constant 1.6 x 10⁴ N/m. How much work does the car do on the spring as it comes to a halt? How much work does the spring do on the car? How much does it compress? Ignore friction.

ok so: .5(1000 kg)(20 m/s)^2 = 2.0 x 10⁵ J is the amount of work the car does on the spring. The spring on the car is the negative of this and the amount of compression is 5 meters. Anyhow after doing that I was wondering if the car was moving at 20 m/s at a constant rate then there would be no net force, and no net work done, so perhaps I was actually calculating it as if at 20 m/s the kinetic energy already implied a certain amount of force to that was there already? For instance what if some object in space that was moving at constant velocity collided with some type of spring that was fastened to some wall. The object has no net force, so what happens? It should stop, but it shouldn't ever compress the spring no matter how large the mass is? Man, I need to get these things straightened out in my head if I want to actually study and apply physics rather than memorize it. [edited by - nervo on March 2, 2004 3:42:48 PM]

SiCrane

Mar 2, 2004 at 2:45 PM

At the very instant of impact, the car does no work on the spring, because there is no net force. However, after that, the spring will compress, and the work will be done on the spring, as there is force: the spring constant times the displacement of the spring.

Nervo

Mar 2, 2004 at 2:50 PM

quote:
Original post by SiCrane
At the very instant of impact, the car does no work on the spring, because there is no net force. However, after that, the spring will compress, and the work will be done on the spring, as there is force: the spring constant times the displacement of the spring.

right, I understand the concept of work being force by displacement. However, if the car was just travelling at constant speed when it hits the spring then how does it compress? If an object is moving at constant speed on a frictionless surface (ignoring air resistance as well) there is no force on it. Hence when it collides with the spring, how can it do work? I know it sounds retarded, but I''m just using the logic that I''ve been learning in physics.

SiCrane

Mar 2, 2004 at 3:01 PM

It compresses by turning kinetic energy from the car into potential energy stored by the spring: the spring supplies the force necessary in this case, which is applied in the direction opposed to the velocity vector of the car.

Cosmic314

Mar 2, 2004 at 3:16 PM

Physics is filled with little riddles like this problem because most calculations are approximations to reality.

For example, in the above problem the implicit statements are:
(1) The car is moving 20 m/s relative to the spring.
(2) The other end of the spring is attached to the Earth, a much much larger object than the car and spring.
(3) The Earth can be treated as a stationary object unaffected by either the car''s or spring''s motion or energy.

In particular the 3rd statement violates the laws of motion. In reality, as the spring compresses the car, the spring in turn is pushing the Earth away. The momentum transfer to the Earth changes it''s velocity, very, very near 0. But the momentum transfer to the spring is very, very near the "ideal" calculation as if the Earth could be treated as a stationary object.

In your wall example in space, the laws of conservation of momentum apply. After all interactions are done, the momentum of the system must be preserved. That is,
massCAR * velocityCar +
massSPRING * velocitySPRING +
massWALL * velocityWALL = momemtum of system (this value is unchanging)

Again that system is an approximation of reality. Einstein says that the momentum of the system is only preserved when you equate mass and energy, using his famous equation E = m*c^2. However, like the Earth example the amount of energy-mass converted is negligible.

The problem becomes complex because not all of the car''s energy will be transferred to the compression of the spring. Some of the energy will be converted into the transveral motion of the spring.

Imagine the wall is very small compared to the car and spring. The car will hit the entire system, the spring will compress a little and the wall will shoot off. The force applied back on the car will be small slowing it down a very, very little. Imagine a tennis ball hanging off a rope in the garage. If the car just "taps" the ball it will have a large increase in velocity.

Imagining the wall to be much larger than the spring and car and you will have a situation just like your original problem. The wall will move a little and the car will rebound.

When the wall/spring is nearly the same mass as the car, the wall and car will net closer velocities depending on the elasticity of the spring.

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