A hockey puck with mass 0.160 kg is at rest at the origin (x = 0) on the horizontal, frictionless surface of the rink. At time t = 0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; she continues to apply this force until t = 2.00s. (a) What are the position and speed of the puck at t = 2.00 s?

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Hey, everyone in this problem, a 25 kg unloaded slate is initially at rest on a horizontal frictionless ice surface. The initial position is at the origin X equals zero. A worker applies a force parallel to the X axis of Nunes A T equals zero on the slave, the forces withdrawn at T equals 2.5 seconds and were asked to determine the speed and position of the sleigh at T equals 2. seconds. So let's just draw a little diagram of what's going on here. We have our sleigh and we're going to draw it as a box on the left hand side. Okay. It is that position X Not is equal to zero and it has an initial speed V not 0m/s. It's initially at rest. So its initial speed is zero. This is the initial case. Now a force gets applied on that box horizontally and it's parallel to the X axis. So the force is going to be applied the side of that box and we've drawn it so that the forces pushing to the right we're going to take to the right as our positive extraction And sometime after our sleigh has been moved further down to the right, we want to know what the position is and what the speed is. And we know that the time here is 2.5 seconds. All right. So we have information about time. We're being asked to find the speed and position. What initially comes to mind is, are you a M or Kinnah Matic equations? So we write out all of our variables for those we have the, not what we know is zero m per second. We have V F which we don't know. But we'd like to find, we have a which we don't know. We have delta X and this is gonna be equal to X final minus X initial, X initial is zero. So this is just equal to X final. And so we'd like to find that quantity T which we know is 2.5. So looking at this, we have two variables that we know to, that we want to find out, but two known variables is not enough to use our equations. We need three. Okay. So we need to find one of these quantities a different way. Well, we are also given information about a force and recall that we can relate the force to the acceleration. Okay. So let's find the acceleration using Newton's law. And then we can come back to find the quantities we're looking for. Recall that Newton's second law tells us that the sum of the forces is equal to the mass times the acceleration. In this case, we only have the one single force that's a force of magnitude 80 Nunes. And we have the mass of the sleigh is 25 kg. Given in the problem the acceleration a that we're trying to find. So we can find a by taking 80 newtons dividing by 25 kg. And this is gonna give us 3.2 m per second squared. Okay. Now, in terms of the unit recall that a Newton is equivalent to a kilogram meter per second squared. Okay. So when we have a Newton and then we divide by kilogram, we get the units of meters per second squared like we want for acceleration. So now if we look back at our U AM equation or Kinnah Matic equation variables, we can fill in that the acceleration is 3.2 m per second squared. Now we have three known variables which means we can solve for the others. So let's start by solving for via the final speed. So we want to choose the equation that doesn't include delta X because we don't know delta X yet, that's gonna be the fault V F is equal to V naught plus A T. So our final speed V F is going to be equal to the initial speed is zero. So that term goes to zero acceleration 3. m per second squared times A time 2. seconds and this is going to give us a VF Of eight m/s. Alright, so there's our final speed Or the speed at T equals 2.5 seconds. We want to do the same to find the final position. So we're gonna choose the Kinnah Matic equation with the final position in it this time. Now we know all of the other variables so we can choose any equation that has delta X. We're going to use the following delta X is equal to V naught T plus one half 80 squirt A delta X again is equivalent to X minus X, not the final position minus the starting position. V, not the initial speed is zero. So this first term goes to zero. We have one half times the acceleration, 3.2 m per second squared times T squared, 2.5 seconds squared. On the left hand side again, Exxon is zero. So we just have X and if we work out the right hand side on our calculator, we are going to get 10 in the unit is meters. We have meters per second squared times seconds squared. So the second squared divide out and we're left with just meters and that is our final position 10 m. So if we go back up to our answer choices, we found that the speed of the sleigh was eight m per second and its position after T equals 2.5 seconds is m. So we have answer choice. Thanks everyone for watching. I hope this video helped see you in the next one.

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Key Concepts

Newton's Second Law of Motion

Kinematic Equations

Initial Conditions and Uniform Motion