A 10 kg box slides 4.0 m down the frictionless ramp shown in F...

Question

A 10 kg box slides 4.0 m down the frictionless ramp shown in FIGURE CP10.73, then collides with a spring whose spring constant is 250 N/m. What is the maximum compression of the spring?

Accepted Answer

Welcome back, everyone. We are making observations about the following system here. Let me go ahead and draw out this incline, make sure that all of my ankle measures are correct for this problem. All right. So here's what we have, we have that a crate that has a mass of kg is released from the top of an incline here. And we are told that the incline makes a angle of 50 degrees with the horizontal here. Now we are told that it slides 3.5 m. So right here, this distance is 3.5 m before hitting a spring. And now this spring has a 320 newtons per meter constant here. And we are tasked with finding what is going to be the greatest compression of our spring that occurs. All right. So how are we going to go about this problem here? First thing that comes to mind when we're relating motion and height and spring constant is the principle of energy conservation here. So we are going to have, we have the following right. We have that our final kinetic energy plus our initial gravitate or sorry, our final gravitational potential energy plus our final spring, potential energy is equal to the initial values of all of those different types of energies here. Now, we actually have formulas for all these terms. So let me go ahead and expand out this equation. We have that this is equal to one half times mass times our final velocity squared plus our mass times gravity times our final height plus one half times our spring constant times our final deformation, which is what we're looking for. Squared, equal to one half times our mass times our initial velocity squared plus M G Y I plus one half times our spring constant times our initial deformation here. All right. Well, what are a couple of things that we know? Well, we know that the brick is going to come to a complete stop here or the crate rather. So our final velocity is going to be zero. Our spring is not origin originally deformed whatsoever. So this is just going to have an initial deformation of zero. What I'm gonna do is I'm gonna say wherever this maximum deformation brings the crate to, I'm going to label that zero m and therefore assign our final height to be zero m. And we also know that the block initially starts at rest, meaning that it has a initial velocity of zero m per second. Now, what about the initial height here? Well, this initial height of our block is just going to be the sign of our angle times the hypotenuse, which we know is going to be 3.5 plus whatever that deformation is that we are trying to find now that we've figured out all those variables here. Let's go ahead and plug in all of those variables to our above equation. And what do we get from this? Well, for our first term, we get zero because our final velocity is zero plus zero because our final height is zero plus one half times our spring constant times our deformation that we are looking for squared equal to zero. Because our initial velocity is zero plus mass times gravity times our initial height, which we said to be 3.5 plus delta R times the sign of 50 plus zero because our original deformation was zero. So what does this work out to be? Well, let's, let's plug in the terms that we know we have that this is equal to one half times our spring constant of 3 20 newtons per meter times. Our desired deformation squared is equal to. Well, I'm gonna distribute the mass, the acceleration due to gravity and the sine 50 here. So we are going to have 25 times 9. times our, let's see here. Uh 3.5 which is the first term in the parentheses, times sine 50 plus our mass 25 times 9.8 times our desired deformation times the sign of 50. All right. So what I'm gonna do now is I'm gonna move all of the terms to one side of the equation. And, and simplifying our terms here numerically, what we get is 1 60 times delta R squared minus 187.68, delta R minus 6 56.88 is equal to zero. What we can observe here at is that this is a quadratic equation. So what we can do is we can apply the quadratic formula to it. So let's go ahead and do that. We have that delta R is equal to negative B which is 87.68 plus or minus the square root of B squared. So we have negative 1 87.68 squared minus four times A which is 1 times C which is negative 656.88. Let me extend that radical here all divided by two times A which is 1 60. What we get from this is that delta R is equal to either 2.7 m or negative 1.52 m. But what we are looking here or looking for here is a positive deformation, the positive distance traveled by the crate once it hits the spring. So a negative value is not going to be physically possible here. So our final answer is 2.7 m which corresponds to our final answer. Choice of D. Thank you all. So much for watching. I hope this video helped. We will see you all in the next one.

A 10 kg box slides 4.0 m down the frictionless ramp shown in FIGURE CP10.73, then collides with a spring whose spring constant is 250 N/m. What is the maximum compression of the spring?

Key Concepts

Conservation of Energy

Gravitational Potential Energy

Spring Constant and Hooke's Law

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