Springs: Inclined ramp, object moving down towards spring

[DPatel304]

Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates weighing 1470 N will move at a speed of 1.80 m/s at the top of a ramp that slopes downward at an angle 25.0. The ramp exerts a kinetic friction force of 540 N on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 7.70 m along the ramp. Once stopped, a crate must not rebound back up the ramp.

Homework Equations

K1 + Ugrav1 + Uelastic1 - Wfriction = K2 + Ugrav2 + Uelastic2

The Attempt at a Solution

K1 = 0.5*150*1.8^2
Ugrav1 = 0
Uelastic1 = 0
Wfriction = (540)(7.7) = 4158

K2 = 0
Ugrav2 = 1470*(-7.7)(sin25) = -4783
Uelastic2 = 0.5k(7.7^2)

Using the above equation, I got k = 29.27. Unfortunately, this is not the correct answer, so I am a little stuck as to what I need to do next.

 

[LowlyPion]

You are designing a delivery ramp for crates containing exercise equipment. The crates weighing 1470 N will move at a speed of 1.80 m/s at the top of a ramp that slopes downward at an angle 25.0. The ramp exerts a kinetic friction force of 540 N on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 7.70 m along the ramp. Once stopped, a crate must not rebound back up the ramp.

I am guessing that the question here is to determine the spring constant?

What is also not determined here is the distance over which the spring will compress. But since they don't want it to rebound from the bottom position then we know at least it must supply a force that satisfies is Hookes F = -kx. And that force would be 1470N*Sin25 + the frictional force from the rollers of 540N preventing it from rolling back up seems to me.

So now you can solve your energy equation

PE loss = 1470*7.7*sin25 ; basically your usual m*g*h
Then there is the work of friction over the distance as you found as well as the initial kinetic energy also found.

And all that you can now resolve into one equation that relates 1/2kx². That and your other equation should allow you to determine x and k right?

 

[nlightner]

It seems like you have set up the problem correctly using the conservation of energy equation. However, there are a few things to consider in order to find the correct answer.

Firstly, it is important to note that the crates will not reach the bottom of the ramp with a constant velocity due to the presence of friction. This means that the kinetic energy at the bottom of the ramp will not be zero, as you have assumed in your calculation for K2.

Secondly, the maximum static friction force of 540 N will only come into play if the crate starts to slide back up the ramp, which we want to avoid. Therefore, we can ignore this value in our calculations.

To find the correct value for the spring constant, you will need to consider the work done by the kinetic friction force on the crate as it moves down the ramp. This work will decrease the kinetic energy of the crate and contribute to the compression of the spring.

So, the correct equation to use would be:

K1 + Ugrav1 + Uelastic1 - Wfriction = K2 + Ugrav2 + Uelastic2

Where Wfriction = (540)(7.7)cos25 = 4469.8 J

Using this equation, you should be able to solve for the spring constant k and find the correct answer. Keep in mind that the crate will not rebound back up the ramp if the spring constant is greater than or equal to the value you calculated above.

I hope this helps and good luck with your calculations!