A horizontal spring with spring constant 85 N/m extends outward f...

Question

A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring?

Accepted Answer

Everyone in this practice problem, we're being asked to calculate the maximum speed of oscillation of a 0.75 kg block attached to a horizontal spring with a spring constant of 400 Newton per meter. The block is going to be pulled towards the right by eight centimeters from its equilibrium position and then released. And we're being asked to calculate the maximum speed of oscillation of the block considering that the motion in the right direction is going to be positive. And assuming that the surface is going to be friction list. The options given for the maximum speed of oscillation is going to be a negative 1.85 m per second. B 1.85 m per second, C 3.41 m per second and D negative 3.41 m per second. So we will use the particle model for the block and I am going to start us off with drawing a visual representation of our system. So uh initially our block is going to be at equilibrium. So our block is going to be 0.75 kg uh just like so and it will be attached to a horizontal spring and will be at equilibrium position. So this is going to be our four position. And at this particular position, um we will model the block as a particle. So at this position, X will equals to zero m just like. So, however, um the block is going to be pulled towards the right and it will then be not at equilibrium position and it will move to a position of X equals to eight centimeter with the spring being compressed just like. So and that will be the after position. So X equals eight centimeters will then equals to eight times 10 to the power of negative two m just like. So it the box is still or the block is still going to be 0.75 kg after. And the work done from this particular scenario can actually be obtained by integrating the force variable. So work will then equals to the integral from zero to the final position X F which is going to be eight times standard of power of negative two m. And the force is then going to be K multiplied by X multiplied by D X for the integral. And the work done W is actually going to equal the change in kinetic energy. So W will equals to delta K. And therefore, uh our delta K can then be calculated by recalling that the kinetic energy is half multiplied by mass multiplied by velocity squared. So delta K is half M V F squared minus half M V I squared. Uh according to the problem statement, the box is going to be initially at rest at its equilibrium. So V I therefore is going to equals to zero. So the right side is then going to equals to zero. So our delta k will then just be equals to half M V F squared just like. So, so combining delta K and W into our W equals delta K equation, we will then get um W equals delta K W S. The integral from zero to X F or K X D X equals to half M V F squared just like. So, so we will start uh substituting every single information that we know into this particular equation right here. So the left side is then going to be the integral from to 8 times 10 to the power of negative two m of K X multiplied by D X. And that will equals to half M V F squared. Then we can substitute the variables which is then going to be the integral from 0 to 8 times 10 to the power of negative two m K is given which is 400 Newton per meter X D X equals to half M. The mass of the box is 0.75 kg. And the V F squared is the uh velocity or the maximum speed of oscillation which is gonna be uh the one thing that is asked in the problem statement. So we wanna recall the integral of C X D X is then going to equals to C over two X squared. So we will utilize this integral formula here into our integration that we have in our expression. And that will give us 400 Newton per meter of X squared divided by two. All of that is going to be evaluated from to 0.8 m or essentially eight times 10 to the power of negative two m. The left side is going to equals to half multiplied by 0.75 kg multiplied by V F squared. And that will essentially give us the left side is then going to be 400 Newton per meter multiplied by 0.8 m squared divided by two equals half multiplied by 0.75 kg multiplied by V F squared. After some rearrangement and calculation P F squared is then going to be 3.41 33 meter squared over second squared. And taking the square root of that V F is then going to equals to the square root of 3.4133 m squared divided by second squared. And that will give us a V F value of 1. m per second. However, since the motion of the block will be first going towards the left because we are pulling the box to the right at first, according to our convention that the things moving to the left side is going to be having a negative value. Then therefore R V F here is then going to be negative 1.85 m per second. So because the motion of the block will be towards the left, hence V F will equals to negative 1.85 m per second. I'm just gonna add a note here for the negative sign because moving two left just like. So, so therefore the correct option is then going to be option A with the maximum oscillation speed of a negative 1.85 m per seconds. And I'll be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topic and I'll be all for this one. Thank you.

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