Spring compression distance from a block sliding down a ramp

In summary, the maximum compression of the spring is 1.46m. At this compression, the box has its maximum velocity.
  • #1
Reverend Lee
5
0
A 10kg box slides 4.0m down a frictionless ramp, then collides with a spring whose spring constant is 250 N/m. (Also, the ramp is angled at 30 degrees).
a). What is the maximum compression of the spring?
b). At what compression of the spring does the box have its maximum velocity?

I started the problem by finding the final velocity of the block at the end of the 4m distance (which is the beginning of the spring) by finding the acceleration ( a=9.8*sin(30) ), then I applied it to the equation Vf ^2= Vi ^2 * 2a(delta x), and received an answer of Vf=6.261 m/s. I used this result for the initial velocity of the block moving down on the spring. After that, I used the equation

.5*m*(Vf^2)+.5*k*(Xf - Xequilibrium)^2 +mgXf =.5m(Vi^2) + .5k(Xi - Xequilibrium)^2 + mgXi

I considered Xi = Xequilibrium = 0; Vf = 0; and Vi =6.261
but got the wrong answer (0.88m). The correct answer is 1.46m. I'm not really sure how to approach part b either.
 
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  • #2
Use energy conservation from the point at which the ball meets the spring to the point of maximal contraction of the spring, it gave the right answer.
 
  • #3
Thats what I tried with the equation, I guess my real problem lies with how I set it up... I went about it like this

(final kinetic energy)+(final elastic energy)+(final gravitational potential energy) = (initial kinetic energy)+(initial elastic energy)+(initial gravitational potential energy)
so I got the equation (each segment is the respective energy)
.5*m*(Vf^2)+.5*k*(Xf - Xequilibrium)^2 +mgXf =.5*m*(Vi^2) + .5*k*(Xi - Xequilibrium)^2 + mgXi

I considered the initial velocity of the block beginning to compress the spring as 6.261, The final velocity of the block as 0 (since it would be motionless at the maximum compression). I also considered the initial length of the spring without being compressed to be 0, as is equilibrium length. That way I could cancel out ".5*m*(Vf^2)", ".5*k*(Xi - Xequilibrium)^2", and "mgXi", which leaves
.5*k*(Xf^2) + mgXf = .5*m*(Vi^2),
giving me a quadratic equation, which I took the positive number for the position. I tried again, but no luck still... my question is, how do I set the equation up correctly?
 
  • #4
You did not account for your inclined plane.

Its going to be pushing up the incline, and I'm pretty certian there is some trig work involved.
 
  • #5
Reverend Lee said:
I considered the initial velocity of the block beginning to compress the spring as 6.261, The final velocity of the block as 0 (since it would be motionless at the maximum compression). I also considered the initial length of the spring without being compressed to be 0, as is equilibrium length. That way I could cancel out ".5*m*(Vf^2)", ".5*k*(Xi - Xequilibrium)^2", and "mgXi", which leaves
.5*k*(Xf^2) + mgXf = .5*m*(Vi^2),
giving me a quadratic equation, which I took the positive number for the position.
Xf is measured along the incline, but potential energy (mgy) uses the change in height (vertical distance). (This is what GoldPheonix was talking about.)

General comment: Nothing wrong with your overall approach (except for that error above), but you could have saved a bit of work by using conservation of energy right from the start. The initial energy is gravitational PE, the final energy spring PE (assuming you set your zero point at the lowest position). No need to calculate the velocity when it hits the spring.
 
  • #6
Ooooh I got it now, thanks much for all of your help folks :biggrin:
 

Related to Spring compression distance from a block sliding down a ramp

1. What is spring compression distance?

Spring compression distance is the distance that a spring is compressed when a force is applied to it. It is a measure of how much the spring has been compressed from its original length.

2. How is spring compression distance related to a block sliding down a ramp?

The spring compression distance is related to a block sliding down a ramp because as the block slides down the ramp, it exerts a force on the spring. This force causes the spring to compress, resulting in a measurable compression distance.

3. What factors affect the spring compression distance?

The factors that affect the spring compression distance include the mass of the block, the angle of the ramp, the coefficient of friction between the block and the ramp, and the stiffness of the spring.

4. How can the spring compression distance be calculated?

The spring compression distance can be calculated using the formula F = kx, where F is the force applied to the spring, k is the spring constant, and x is the distance the spring is compressed. This formula can be rearranged to solve for x, which gives the spring compression distance.

5. Why is the spring compression distance important in this scenario?

The spring compression distance is important because it allows us to measure the amount of potential energy stored in the spring. This potential energy is then converted into kinetic energy as the block slides down the ramp, and the spring decompresses. This relationship between potential and kinetic energy is an important concept in physics and can be applied to various real-world scenarios.

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