FIGURE EX6.10 shows the force acting on a 2.0 kg object as it ...

Question

FIGURE EX6.10 shows the force acting on a 2.0 kg object as it moves along the x-axis. The object is at rest at the origin at t = 0 s. What are its acceleration and velocity at t = 6 s?

Accepted Answer

Hey, everyone in this problem, we have a six kg box initially stationary at X equals zero m subject to a horizontal variable force, which is shown in the figure we're asked to calculate the acceleration of the box and its velocity after five seconds. Now, the diagram or the figure we shown has the time t in seconds along the X axis and the force FX in newtons along the y axis from zero seconds to two seconds, the force is constant at negative three newtons from to 4 seconds. Again, the force is constant but this time at six newtons and from 4 to 5 seconds, that force once again is constant at zero newtons. Now we're given four different answer choices. OK? And they just differ in what acceleration and velocity they have at that time of five seconds. Now, let's start with the first part of this question that's asking for the acceleration. We're given a force on the graph. And let's recall that we can relate forces to acceleration through Newton's second law. So we have that this force FX is going to be equal to the mass M multiplied by the acceleration. A X well A tea was five seconds. The force fx according to our diagram is that zero nuance, OK. It's lying along that X axis. And so we get that zero newtons is equal to the mass of six kg multiplied by the acceleration A X. And so the acceleration A X must be zero at this time point. So the acceleration of five seconds is zero m per second squared. What that means is we can already eliminate two answer choices. Option A and B have that acceleration of zero m per second squared. But option C and D have accelerations of negative and positive 0.5 m per second squared respectively. And so we can eliminate both of those options. Now, let's work on velocity. OK. That's gonna help us figure out whether we have answer A or answer B. Now we need to break this up into each part and we need to start at T equals zero and work our way up to T equals five in order to figure out what's going on with the velocity. So from zero seconds to two seconds, what do we have? Well, we know that the initial velocity V knot is going to be equal to zero m per second. And we're told that this box is initially stationary so that initial velocity is zero, we don't know what the final velocity at two seconds is. We know the acceleration or we can find it out using our diagram, we don't know how far the box travels and we know that the time taken is two seconds. OK? We're going from zero seconds to two seconds. That's a two second interval. All right. Now, what we need to find from this time point is the final philosophy. OK. The final velocity at two seconds is going to be the initial velocity for the next segment. OK. Then we can do the same thing with that segment. Figure out the final velocity which is gonna be the initial velocity of that final segment. So we're working our way up to that final velocity. So we wanna calculate in V F here and again, the acceleration we can calculate using that graph. So just like we did above, we know that the force FX is equal to the mass multiplied by the acceleration A X. Let's go to our diagram and from zero seconds to two seconds, that force is negative three nodes. And so we have that negative three newtons is equal to the mass of six kg multiplied by the acceleration A X dividing by six kg to get our acceleration. Ok. Recall that newtons are equivalent to kilogram meter per second squared. So when we divide by kilogram, we're left with just meters per second squared, which is the unit we want for acceleration and we get negative 0.5 m per second squared. So we can add that into our information. Negative 0.5 m per second squared. And now we have three known values B, not A and T, we can use those to calculate V F. So we're gonna choose the kinematic equation that does not include delta X and we don't have information about it and that's not what we're looking for. And that equation is going to be V F is equal to V plus a multiplied by T our initial velocity is zero m per second. OK. Plus the acceleration, negative 0.5 m per second squared multiplied by the time of two seconds. And this gives us a final velocity of negative one m per se. And again, that's for this interval from 0 to 2 seconds. So we know the final velocity from 0 to 2 seconds. Now let's work on the next interval. I'm just drawing a dotted line to separate that interval. Now, the next interval starts at two seconds and goes up to four seconds. OK? We're just picking the intervals where we have different forces in that graph. Now, the initial velocity for this interval that starts at two seconds is gonna be the final velocity from the previous interval that ended at two seconds. And that's gonna be negative one m per second. Again, we wanna find the final velocity so that we can use it as the initial velocity in our next segment. The acceleration we're gonna calculate in the exact same way we did before. We don't know the distance traveled. And again, the time is two seconds, we go from two seconds to four seconds. OK. That's a change in time of two seconds. Now, calculating our acceleration again, F of X is equal to MA X going to our diagram, the force from 2 to 4 seconds is six nodes. So we get that six newtons is equal to six kg multiplied by A X. And so our acceleration for this segment is going to be one m per second squared. OK. This is a positive acceleration. This time before we started stationary, we had a negative acceleration and we ended up with a negative velocity. Now we're starting with a negative velocity but we have a positive acceleration so we can add that one m per second squared in that information for acceleration. Now we have three notes V not A T, we wanna calculate V F same as before V F is equal to V nine plus 80. This is equal to negative one m per second plus the acceleration of one m per second squared multiplied by the time of two seconds. OK? One multiplied by two gives us two negative one plus two. This gives us a final velocity of one m per second. And again, this is from two seconds to four seconds. We have this final velocity. Now we have one more segment to look at. Can I remember we're looking for that velocity after five seconds, we've made it up to four seconds, we have one more segment to look at. So again, I'm gonna extend this dotted line just so that we can separate the information. So we're not mixing it up between segments. And now we're looking oops from four seconds to five seconds. Now, the initial speed in this segment is gonna be the speed at four seconds. Which was the speed or velocity, I should say sorry at the end of the last segment. So our initial velocity is one m per second. The final velocity is what we are trying to calculate. We know the acceleration because we calculated it at the very, very top to answer the first part of this problem. Ok. We know that that acceleration is zero m per second squared. We don't know the distance traveled should be delta X and we know that the time here is just one sec. So we see an acceleration of zero. We know that that means our velocity is not changing. So we expect that final velocity to be one m per second. A we have a constant speed and so V F is just going to be equal to one m per second, ok? And if you didn't notice that right off the bat, you can still use the equation we had before with the acceleration equal to zero, you get that V F is equal to B not and that is that final velocity, the velocity at five seconds that we were looking for. So we started this problem by finding the acceleration using our graph and new second law, we found that the acceleration was zero m per second squared at five seconds. Ok. So we eliminated option C and D then we calculated the velocity of one m per second at five seconds. And so the correct answer here is going to be option B Thanks everyone for watching. I hope this video helped see you in the next one.

FIGURE EX6.10 shows the force acting on a 2.0 kg object as it moves along the x-axis. The object is at rest at the origin at t = 0 s. What are its acceleration and velocity at t = 6 s?

Key Concepts

Newton's Second Law of Motion

Kinematics

Graph Interpretation

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