# Physics spring/crate hits another spring problem

• MattNguyen
In summary: That's a very different physical situation.In summary, a crate with a mass of 0.29 kg is set against a compressed spring with a spring constant of 581 N/m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of 246 N/m. In part a, it is asked how far the second spring will compress when the crate runs into it, and the answer is 0.1537 meters. In part b, the crate's velocity when it strikes the second spring is being sought and the equation .5mv^2+.5kx=2.905 is used, though it is unclear what the value of x should be. Finally, in part c
MattNguyen

## Homework Statement

A crate of mass m = 0.29 kg is set against a spring with a spring constant of k1 = 581 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 246 N/m.

(a) How far, d2 in meters, will the second spring compress when the crate runs into it?

(b) How fast, v in meters per second, will the crate be moving when it strikes the second spring?

(c) Now assume that the surface is rough (that is, not frictionless). You perform the experiment and observe that the second spring only compresses a distance d2/2. How much energy, in joules, was lost to friction?

## Homework Equations

W=change in mech energy=Kf+Uf-Ki-Ui

## The Attempt at a Solution

[/B]
I already calculated part a to be 0.1537. I just don't know how/if it relates to part b.

So far I have .5mv^2+.5kx=2.905 for part b. I'm just unsure what x is supposed to be or if I even set it up correctly.

MattNguyen said:
I'm just unsure what x is supposed to be
So am I. It's your equation.
First, I assume you meant x2. Secondly, at the point in time indicated for part b, neither spring is under compression.

Yeah, that's what I meant. The first spring is stretched, though, so I'm assuming it has potential energy.

MattNguyen said:
Yeah, that's what I meant. The first spring is stretched, though, so I'm assuming it has potential energy.
It does not say it is attached to the first spring, merely pushed against it.

## 5 Most Frequently Asked Questions about "Physics spring/crate hits another spring problem"

1. What is the concept behind a spring/crate hitting another spring in physics?

The concept behind this problem is the conservation of energy and momentum. When the two springs collide, they exchange energy and momentum, resulting in a change in their individual properties.

2. How do you calculate the velocity of the springs after the collision?

The velocity of the springs after the collision can be calculated using the laws of conservation of energy and momentum. The total energy and momentum before the collision should be equal to the total energy and momentum after the collision.

3. Is there a difference in the collision between two springs with different spring constants?

Yes, the spring constant affects the velocity and displacement of the springs after the collision. A higher spring constant results in a greater velocity and displacement, while a lower spring constant results in a lower velocity and displacement.

4. Can the spring/crate hitting another spring problem be solved using only one-dimensional motion equations?

Yes, this problem can be solved using one-dimensional motion equations as long as the motion is constrained to a single axis.

5. How does the mass of the crate affect the collision between the two springs?

The mass of the crate affects the momentum of the system. A heavier crate will have a greater momentum, resulting in a larger change in the velocity of the springs after the collision.

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