In a physics lab experiment, a compressed spring launches a 20...

Question

In a physics lab experiment, a compressed spring launches a 20 g metal ball at a 30° angle. Compressing the spring 20 cm causes the ball to hit the floor 1.5 m below the point at which it leaves the spring after traveling 5.0 m horizontally. What is the spring constant?

Accepted Answer

Hey, everyone in this problem, we have an experimental procedure for determining the force constant of a spring that uses a compressed spring to shoot a 50 g cube at 40 degrees above the horizontal spring compressed by 25 centimeter projects. The cube from a flat bench raised by 1 m above the ground. The cube covers a horizontal distance of 4 m before landing on the ground. And we are asked to determine the springs force constant. We're given four answer choices all in newtons per meter. Option A 5.69 option B 1.42 option C 6.45 and option D 27.1. So what we're gonna do is start by drawing this up. So we have our spring and it is at an angle of 40 degrees and that spring is gonna shoot a cube and the spring is gonna compress down by 25 centimeters. OK. So we can imagine squishing this all the way down. So that cube is in line with the bench, OK? And that spring or that cube is gonna go on some projectile motion path afterwards and land on the ground which is 1 m below the bench and it's gonna land 4 m away horizontally from where it started. And so that's what we have going on. We're gonna label up into the right as our positive directions and we'll keep going with this problem. So what we're looking for is the spring's force constant. And what we're gonna start with is by thinking about conservation of mechanical energy, we can assume that we have no net forces acting on the system. And so the mechanical energy will be conserved. The reason we want to think about mechanical energy is because we're gonna have a spring potential energy component there that will allow us to introduce this spring force constant that we're looking for into our equation. So let's think about the compression of this. OK. So what we can say is that the mechanical energy when the spring is compressed, we'll use subscript C, it's gonna be equal to the mechanical energy of the spring of the, of the system story initially. And when I say initially, what I mean is at the beginning of the projectile motion. OK. So after the spring has returned to its resting place and the cube has started its projectile motion, that's gonna be our initial time. So we have our compressed spring and then the initial portion of the projectile motion. All right. So what makes up our mechanical energy? We have kinetic energy, gravitational potential energy and spring potential energy in this case. So we're gonna have EKC plus EGC plus EU C. What you cannot not plus EG not plus E you know now when this spring is at maximum compression, the cube is not moving. And so the kinetic energy is going to be zero. So what we can also do is assume that when this spring is compressed that that is gonna be considered height of zero. OK. At height of zero, the gravitational potential energy will be zero. OK. So all we have when the spring is compressed is that spring potential energy and that's gonna be converted into these other forms of energy. Now, what we won't have in the initial case, once this, the cube has started, its projectile motion is spring potential energy that spring will return to its resting place. There'll be no more spring potential energy. And so eu not will be easier. So we're converting all of that spring potential energy into kinetic and gravitational potential energy. OK. Let's go ahead and write out these energies in their full form. OK. So what we're gonna have is the spring potential energy we call is one half kx squared. So we get one half K XC squared will be equal to, we have kinetic energy one half MV squared. So one half MV knot squared plus the gravitational potential energy MG H nothing. Now, we know how much the spring compresses by. We're told that in the problem, we know the mass of our cube, we know the acceleration due to gravity, we can actually find H knot. So if we think about labeling our diagram H knot is going to be the distance between the horizontal where our cube is remove from that spring. OK. So once that spring has returned to its resting position and the cube is released from it, that is going to be our height for gravitational potential energy at the beginning of that projectile motion. So if we think of this little triangle, we've just made, we have each knot on the opposite side and we have an angle of 40 degrees and we have a hypotenuse of 25 centimeters. So we can write that H knott will be equal to 25 centimeters or 0.25 m multiplied by sign of 40 degrees. OK. So the last thing we need to figure out is what is this V no value? So let's go ahead and think about this V not value. And in order to do that, we're gonna look at our kinematics or U AM equations. And so we're gonna start by thinking about all of the variables we have acting, we're gonna look in both the horizontal direction and the vertical direction. So in the horizontal direction, we can assume we have no air resistance. So the acceleration will be zero, the speed will be constant. OK. So our speed in the horizontal direction which we'll call VX, this is just going to be the initial speed V knot that the cube leaves the spring with multiplied by cosine of 40 degrees. OK. To get the X component, we know that the horizontal distance traveled delta X is 4 m. And we don't know how long this takes. We can say that TX is equal to some time T and so if we want to put this into an equation, we can say that the speed VX is equal to delta X divided by delta T substituting in our values VNO cosine of 40 degrees is equal to 4 m divided by T. And we're actually gonna rearrange and solve for T. And we're gonna do that because this time T is gonna be the same in the horizontal direction as it is for the vertical direction that will allow us to substitute that in and solve for the unknown that we want, which is Von. And so we can write that T is equal to 4 m divided by V not cosine of 40 degrees. All right, let's move to the vertical direction. And right now we have an equation with two unknown values. If we look in the horizontal or sorry, the vertical direction, we're gonna get another equation with those two unknowns, two equations, two unknowns will be able to solve. So in the vertical direction, we have that the initial velocity we write as that speed, we not multiplied by sign of 40 degrees. OK. So this time taking that vertical component, we know about the final velocity and we know that our displacement delta Y is gonna be negative 1 m and we're going down to the ground. So down that negative 1 m, our acceleration is gonna be the acceleration due to gravity negative 9.8 m per second squared. And the time this takes ty is the exact same time as in the extraction which we just called T, OK, whether we're looking in the vertical direction or the horizontal direction, the cube lands on the ground at one time. All right. So if we're thinking about what U AM equation to use, we're gonna choose the one with the four variables we've written. Now, those have information that we are interested in. We can ignore the equation with the final velocity. So we get delta Y is equal to V not Y multiplied by T plus one half multiplied by A Y multiplied by T squared. Substituting in our values negative 1 m is equal to V not dying of 40 degrees, multiplied by T plus one half multiplied by negative 9.8 m per second squared multiplied by T square. Now, we again have an equation with two unknown values. What we're gonna do is substitute in the equation we found for t earlier, we're gonna call that equation star. So T is equal to 4 m divided by VNO cosine of 40 degrees. We're gonna substitute that into our equation, we get negative 1 m is equal to V knot sign of 40 degrees multiplied by 4 m divided by V, not cosine of 40 degrees minus 4.9 m per second squared multiplied by T square 4 m divided by V knot, cosine of 40 degrees. All scored. We can go ahead and simplify. So we have V knott divided by V knot that will divide out on the right hand side, we have sin of 40 degrees divided by cosine of 40 degrees. That's just equal to tangent of 40 degrees. So we're gonna go ahead and move our term on the right hand side to the left side. We're gonna move our 1 m to the right hand side. OK. What we're gonna have is that 4 m multiplied by a tangent of 40 degrees plus 1 m is equal to 4.9 meters per second squared, multiplied by 16 m squared, divided by VNO squared cosine squared of 40 degrees. Now, we wanna solve for VOT and so we're gonna multiply both sides by V not squared. Then we can divide by the rest of the left hand side. V knot squared will be equal to 4.9 m per second squared, multiplied by 16 m squared, all divided by cosine squared of 40 degrees multiplied by 4 m tangent of 40 degrees plus 1 m. OK? So we can work all of this out on our calculator. We're gonna take the square root. And we're gonna use the positive value because we're going into our energy equation. We're using this value in our energy equation. We care about the speed. We don't care about the velocity of the direction doesn't matter in this case. So we're gonna take the positive value and we get that V knot is equal to approximately 5.53784 m per second. OK. So we have this speed, the speed that the cube leaves the spring with, we can get back to our conservation of energy equipment. So we come back to this equation. We have everything we need now to solve it. And so what we're gonna have is one half multiplied by K that value we're looking for that's from constant multiplied by the compression K 25 centimeters, 0.25 m squared. It's gonna be equal to one half multiplied by the mass. And we have 50 g converting to kilograms, 0.05 kg multiplied by V not squared. That speed we just found. So five 0.53784 m per second. That is awkward. And then we're gonna add that gravitational potential energy. So the mass again, 0.05 kg multiplied by 9.8 m per second squared, multiplied by that initial height, we found 0.25 m multiplied by sign of 40 degrees. And while we work this out and on the left hand side if we simplify, we get 0.03125 m squared multiplied by K. And on the right hand side, yeah, that this is equal to 0.845433 Newton meters. We can divide both sides by our 0.03125 m squared. To get that this spring constant K is going to be equal to 27.0539 newtons per meter. Ok. So this is the spring forest constant we were looking for, ok. We need to use our conservation of mechanical energy as well as some information about kinematics. In order to solve this problem. If we take a look at our answer choices and we round to three significant digits, we can see that the correct answer corresponds with option D. Thanks everyone for watching. I hope this video helps see you in the next one.

In a physics lab experiment, a compressed spring launches a 20 g metal ball at a 30° angle. Compressing the spring 20 cm causes the ball to hit the floor 1.5 m below the point at which it leaves the spring after traveling 5.0 m horizontally. What is the spring constant?

Key Concepts

Projectile Motion

Spring Constant (Hooke's Law)

Energy Conservation

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