A hockey puck with mass 0.160 kg is at rest at the origin (x = 0)...

Question

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. (b) If the same force is again applied at t = 5.00 s, what are the position and speed of the puck at t = 7.00 s?

Accepted Answer

Hey everyone. So today we're dealing with the problem regarding cinematics. So we're being told that a .875 kg truck is initially at rest on a horizontal frictionless surface. Initial position is at a point X. On the origin where X is equal to zero. My bad and a child applies a force that is parallel to the X axis equal to 1.5 newtons at time is equal to zero on the toy. That forces withdrawn at time 1.8 seconds. And is once again applied the time, 4.2 seconds. With this for being us determined the position of the toy at 6. seconds. So we're going to need to draw out a few things. So let us draw our horizontal plane right? Where this is our starting position, X is equal to zero. Now we don't know the positions here, right? We only know time intervals, Right? We have time is equal to zero And initial velocity is equal to zero and where this is initial velocity in the X direction. Right? Then we have a period of 1.8 seconds. So this is at 1.8 seconds. This is at 4.2 seconds. And right here we have our final .6.50 seconds. What we're trying to what the position at which we're trying to find the toy at at that time, at the time, at which we're trying to find the position of the toy. My bad. So with this here we can go ahead and do some quick things. So we know that in this first interval in this first interval there's a force applied. So force through 1.8 seconds. Then during this Uh 2.4 2nd period, no force During this 2.4 2nd period. And then the 2.3 2nd period from 4.2 to 6.5 seconds. Yet again has another force force applied again app again For 2.3 seconds. Now the forces parallel to the X axis, meaning that if we are to have a horizontal plane, we have a horizontal plane like so and we have our little truck here, then the force being applied is parallel to the X axis. So we're only dealing with the horizontal motion. So we only know the time values that we're needing to find out and we only know time. We only, oh well rather we only know mass in the force being applied at the initial start. The force being applied here. So this is, we'll say this is part of our force body diagram this is as well, but the next part that we need to do when dealing with cinematics or just most forced problems, is to factor in Newton's second law. Right? And Newton's second law states that the sum of all the forces is equal to mass, times acceleration. Now in this case we know the force that is being applied At the initial time at Time zero. So we're considering right now for the sake of this problem at time is equal to zero, right? So at time is equal to zero F. is equal to 1.5 newtons and were given the mass right here. So let's plug this in. We have 1.5 newtons is equal to 0.875 Kg. multiplied by a. So A is equal to 1.5 and Divided by 0. Kg Is equal to 1. meters per second squared. So this is the acceleration experience by the excuse me by the truck. Initially as the as the child applies the force that is parallel to the X axis from here. We now have three quantities needed for uniformly accelerated in motion problems. Right? We have the time, we have the time that the forces applied, we have the acceleration and we have the initial velocity. Now we're trying to find both the displacement as well as the final velocity. But let's solve for final velocity first as it is a little easier to get to and let me just scroll down a bit. So we have a little more space. So step three, We get that time is equal to 1.80 from rest 1.8 seconds. Initial velocity is equal to zero m per second because again we are from rest acceleration in the X direction, that's what we're focused on is 1.71 m per second squared and we're trying to solve for both the position as well as the final velocity. Oops not why should be F. So let's solve for the final velocity first and I'll do this in blue. We can use a equation that does not in fact use a Or that does not use displacement. So when I say initial and final right now I am talking initial to final. This 1.8 2nd period. So this is initial and final. The final is equal to the initial. Excuse me Plus substituting in our values and solving for the final velocity we get 1. My bad. We get zero plus 1.7, m/s squared Into 1. seconds Which gives us an answer of 3.08 meters per second. So with the final velocity now we can go ahead and actually solve for the position after 1.8 seconds. Right? So recall. And we'll say that this is Uh delta x 1.8 Right? Delta eight at 1.8 seconds. But it will be equal to the initial velocity multiplied by time plus half the acceleration. Oops, I will do half the acceleration into times square. So solving this will get zero plus one half Into 1.7, 1 into 1.8 zero squared Which gives us an answer of 1. meters. Now we need to do this again for the other two time intervals and the reason why is because the acceleration stops for a little while and then starts again, right? Because the boy stops applying the force so the acceleration therefore becomes zero during that time and then the forces applied yet again. So with that in mind let's go ahead and take a look at the next two times. I'm gonna scroll down a bit. But just remember that we have our values. So now we have it from Time 1.8 - 4.2. So the time that is being traveled and again we are talking about this time interval. This time interval from here to here from 1.8 seconds to 4.2. So during that time interval Where it is 2.4 seconds of time, right? The acceleration is zero because there's no force being applied. zero m per second squared. The initial velocity will be equal to the final velocity because again there is zero acceleration which means the velocity stays constant, not V. Y is the final. This will be 3.08 m/s which is the velocity calculated here and now we need to find the change in position and we'll say this change in position. 2.4 seconds for the 2.4 2nd interval, 2.4 seconds is what we're looking for. This is our target. Now let me write this in rent. We can use the same formula. We have delta X. 2.4 seconds is equal to P, not T plus half a T square. And recall in the initial and final. Now this case, this represents our new V initial. This represents our new the final because that is the time interval that we are concerned with anyways Subsequuting in our values that we have. We get that it is 3.08 m/s Into 2.4 zero seconds. Oops, 2.40 seconds. Our acceleration is zero, which means that the last term becomes zero by default. And we get that this is 7.39 m. So at the end of that 2.4 2nd Interval, The uh small truck has now reached 7.39 m from the origin. So it has gone. You can also rephrase this as plus 7.39 m from X equals zero. Again we do this for the last set of variables because that will tell us the distance traveled. Oh sorry, I think I explained it wrong By wording, this is the distance traveled during this specific time interval. But for the 2.4 seconds, this is the distance travel During the first time interval from to 1.8 seconds. This is this has traveled from 1.8 to 4.2 seconds. Now we find the distance traveled from the uh 4.2 to 6.5 2nd interval which is 2.3 seconds. Our acceleration will be the same as what was applied during the first interval. From 0.21 or from 0 to 1.8. So it'll be 1.71 m per second squared again the same as the first time interval. The initial velocity will be Excuse me, 3.08 m/s because the velocity will remain the same as it was at the end of the last time period due to there being no friction on the horizontal surface. We're searching for the Position or the chains in position during the 2.3 2nd interval. But to do that, we need to also find the final velocity. The final velocity at the end of this time period. We'll use green once again recall that when I say the initial and the final. Now that we've gone to the next time period, this is now our new the initial and this will be our the final for this specific time period. So we have to do what we did for the first time period. We know that the the final v not plus a. T. Which means it is 3.08 plus 1.71 into 2.3 Which gives us an answer of 7.01 m/s. So that is the final velocity of this time period. With this, we can go ahead and solve for the displacement over this period of time. So the change in position During the 2.3 2nd time interval is equal to v, not T plus one half a t squared, oops, 80 squid Subsequuting in our values. That is 3.08 in two, 2.3 plus one half Into 1.7, 1 into 2.3 zero squared. Which gives us a final answer of 11.6 m. So now we have all the information that we need to answer this question At the end of 6.5 seconds, the total change in displacement. The total change will be equal to Changing displacement for the 1.8 seconds plus the change in displacement For the 2.44 seconds plus the change in displacement Or changing position rather for the 2.3 seconds. So this will be 1.5 let me just double check, yep, That'll be 1.54 m plus 7. m plus 11.6 m. So the total change will therefore be equal to 20.5 m Or if you look back at our answer choices, the total change in position for the Excuse me. The change in position as well as the speed will be answer choice, a 7.1 m per 2nd, 7.0 or it will be 7.1 m per second for the speed at 6.5 or 6.5 m per second and the change in total the total change in position Is 20.5 m. I hope this helps, and I look forward to seeing you all in the next one.

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