A 2.0 kg object is moving to the right with a speed of 1.0 m/s wh...

Question

A 2.0 kg object is moving to the right with a speed of 1.0 m/s when it experiences the force shown in FIGURE EX11.8. What are the object's speed and direction after the force ends?

Accepted Answer

Hey, everyone in this problem, a block with a mass of 2.4 kg is initially moving to the right with a velocity of 2 m per second. It's then subjected to a force that varies with time as shown in the figure below and works to determine the magnitude and direction of the block's velocity after the force stops acting on it. So we're getting our diagram here, we're showing the force say the X component of the force in newtons on the Y axis, the time seconds on the X axis. OK. And the force starts at zero newtons from zero seconds to two seconds. OK? It then decreases linearly from zero newtons to negative four newtons from 2 to 6 seconds. Then it increases linearly from six seconds to 10 seconds. Pay back up to zero noons and then remains at zero noons for the remainder of the time. Now we have four answer choices here all in meters per second. Option A negative 5.54 option B negative 3.21 option C negative 4.67 and option D negative 5.48. So what we have here is this force time curve we're looking at velocities. So let's think about our impulse momentum theory. OK. We have this force exerted over a set time interval is we can think of this as an impulse. And let's recall that the change in momentum delta P is gonna be equal to the impulse, which we can write as J. All right. So we have our momentum or our change of momentum delta P equal to the impulse. And recall that the impulse is gonna be equal to that net force multiplied by the change in time delta T. And when we're given an FT curve, this is just gonna be equal to the area under that curve for a change of momentum is just gonna be equal to the area under the FT. All right. So now we have a way to relate the momentum which we know contains information with the velocity to this force in this time interval that were given in our diagram. So let's write out what we know about the momentum, what we know about this area and see if we have all of the information we need to solve for this velocity after the force stops acting. OK. So we change of momentum. So is this gonna be the momentum P will say at 10 seconds? P 10 minus the momentum at two seconds? OK. So P 10 is the momentum at 10 seconds after that force stops P two is the initial momentum when we, when that force is first applied, OK. And this is gonna be equal to the area under that curve. Now, the area we can tell that this is a triangle. And so this entire area is gonna be given by one half multiplied by the base of the triangle multiplied by the height of the triangle. Now, momentum recall is equal to the mass multiplied by the velocity. And so we get the mass multiplied by the velocity at 10 seconds minus the mass multiplied by the velocity at two seconds is equal to one half, multiplied by the base multiplied by the. And what we're looking for is this V 10. OK. The velocity after the force stops acting on it at that 12th mark. We know the mass of our block. We can find the base in the height of our triangle. The only other thing we need is the velocity V two. Now we're told the initial velocity at zero seconds. OK? Is 2 m per second. What we can see is that from zero seconds to two seconds, our force is zero. They now recall according to Newton's law, that the force is equal to the mass multiplied by the acceleration. OK. So if the force is equal to zero Newtons, that tells us that the acceleration must be equal to 0 m per second squared. OK. So if we start at 2 m per second, we have an acceleration of zero, then our velocity at V two is gonna be the exact same as it was at zero seconds and we're gonna have 2 m per second. So let's go ahead and substitute all this information in now. So we have our mass of 2.4 kg multiplied by B 10. This velocity we're trying to find minus our mass 2.4 kg multiplied by the velocity at two seconds, which we've just found to be 2 m per second. And all of this is gonna be equal to one half multiplied by the base of our triangle. OK? Now, the base of our triangle, we go from two seconds up to 10 seconds. So that base is gonna be eight seconds and for the height of our triangle and notice that we're going in the negative direction, OK? Our triangle is going downwards and it has a height of negative for. And that sign is really, really important to getting the correct answer here. All right. So we've written all of this out. Now, we have an equation with only one unknown this V 10 value that we're gonna find. And let's go ahead and simplify. So on the left hand side, we have 2.4 kg multiplied by V 10, we're gonna move our 2.4 kg times 2 m per second to the right hand side by adding it on the right hand side, we have one half multiplied by eight seconds multiplied by negative or newtons. So this is gonna give us negative 16 kg meters per second. And then we're adding that 4.8 kg meters per second that we moved from the left hand side. OK. Simplifying on the right hand side, our equation becomes 2.4 kg multiplied by V is equal to negative 11.2 kg meters per second. And finally divided by 2.4 kg, we get the V 10 is going to be equal to negative 4.66 repeated meters per second. And that is our final velocity. And you'll notice that this velocity has a negative in front of it. Ok? So that indicates the direction. So initially our velocity was positive and our block was moving to the right now. Our velocity is negative which tells us that the block is going to be moving to the left. OK? So we apply this negative force. It was enough to change the direction of our block. If we round to two decimal places, we can see that the correct answer here is going to be option C negative 4.67 m per second. Thanks everyone for watching. I hope this video helped you in the next one.

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A 2.0 kg object is moving to the right with a speed of 1.0 m/s when it experiences the force shown in FIGURE EX11.8. What are the object's speed and direction after the force ends?

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