physics: spring constants (solved)

A spring of length 0.80m rests along a frictionless 30-degree incline. A 2.0kg mass, at rest, at the end of the spring compresses the spring by 0.10m. (a) Determine the spring constant k. (b) The mass is pushed down, compressing the string an additional 0.60m, and then released. If the incline is 2.0m long, determine how far beyond the rightmost edge of the incline the mass lands.

I've attached a little drawing of the problem that i drew quickly just so you can visualize the main parts.

*To refresh your memory if you've forgotten the basics or something:

PE(string) = 1/2kx^2 -- x = distance, k = spring constant
PE(gravitational) = mgh -- m = mass, g = gravity (9.8m/s^2), h = height above reference point
KE = 1/2mv^2 -- m = mass, v = velocity

This sounds a lot like homework... Is there something specificly you are having problems with?

For part (a) you need to know the spring's natural length (0.8) and it's extension or compression (0.1). When it's at equilibrium (i.e. when the block is resting with the spring 0.7 long), the tension provided by the spring points up the slope and the weight of the block points downwards. If you resolve the components of these forces along the slope you will get:

T - 2gsin30 = 0

Since T = 0.1k, you can stick that in to get k = 10g.

For part (b) you need to work out how fast it's going when it leaves the top of the spring, so from the x=0.7 to the x=0 positions:

EPE loss = GPE gain + KE gain

5g(0.7)² = 0.8sin30 + v²

Assuming g = 9.8 this gives you v = 4.859 (not a nice number). You then need to work out how fast it's going when it leaves the top of the ramp, so from when it leaves the spring to when it leaves the ramp:

GPE gain = KE loss

2g x 1.2sin30 = (4.859² - v²)

This gives you v = 3.442 (also not a nice number). Anyways, from this point onwards the block's motion is that of a projectile moving freely under gravity (I'm assuming you don't want air resistance because it gets a lot more complicated).

The horizontal component of v as the block leaves the ramp is 3.442cos30 = 2.981, and the vertical component is 3.442sin30 = 1.721. Use s = ut + at²/2 vertically upwards to get the time taken for it to hit the ground:

-2sin30 = 1.721t - 4.9t², giving t = 0.660.

Finally use s = ut + at²/2 horizontally to get the distance travelled in that time:

d = 2.981*0.660 = 1.968

The numbers aren't nice so I've probably made a silly error somewhere in there but that's the quickest way I can think of to do it.

Quote:

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Originally posted by twanvl
This sounds a lot like homework... Is there something specificly you are having problems with?

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Yes, in fact it is homework. It's one of those physics problems in the back of the chapter that's really hard. I had a problem with the entire problem at first...well, basically setting it up, but I finally figured it out after staring at it for a long time.

For (a) you use the following equation:

F(string) = kx = F(gravity in x direction) = mgsin(x)

For (b) you use potential spring energy, gravitational potential energy, and kinetic energy in the following equation to get the velocity when it leaves the ramp:

PE(spring) = 1/2kx^2 = 1/2mv^2 + sin(x) * mgh

Once you have the acceleration, use the following to find the time it will travel in the air:

sin(x) * lenght of ramp = sin(x) * vt - 1/2gt^2

When you have time and velocity use the following equation to get the distance it will travel:

x = v * cos(x) * t

The final result is 1.2m