Good morning. Billy, please read the problem and Bobby please translate.
♪ Flipping Physics ♪ A 5.0 kg mass experiences a net force causing
the object to go from an initial velocity of 4.0 i minus 3.0 j meters per second to a final velocity
of 2.0 i plus 6.0 j meters per second. What is the net work acting on the mass during this event?
… Mass equals 5.0 kg. Velocity initial equals 4.0 i minus 3.0 j meters per second.
Velocity final equals 2.0 i plus 6.0 j meters per second. And net work equals question mark.
Thank you both. Bo, please begin solving the problem.
Well, we have 2 equations for work. Work equals the dot product of the force doing the work
and the displacement of the object. We use that equation when the force is constant. … The other
equation is work equals the integral from initial to final position of the force in the x-direction
with respect to x. We use that equation generally when the force is not
constant. So, is the force causing this work constant or not constant?
… Uh, we do not have the force or the displacement.
Yeah, and neither of those equations have mass or velocity in them.
Right. … Actually,
we are not going to use either of those equations. We are going to use the Work Energy Theorem.
Oh yeah, duh. Why “duh”?
It’s in the video title. What video title?
The one right above your head. I do not see it.
It’s right there. Okay. Yeah. So,
we need to remember that we can use what is typically called the Work Energy Theorem.
I prefer to call it the Net Work equals Change in Kinetic Energy Theorem:
(I still do not see the video title. What video are we talking about anyway?)
I am sorry Billy. It’ll all make sense to you someday.
… Alright, I prefer to call this equation the Net Work equals Change in Kinetic Energy Theorem
rather than the nebulous Work Energy Theorem because too many of my students forget it
involves the NET work and the CHANGE in kinetic energy. … Bobby, please solve the problem.
Okay, well, we know net work equals change in kinetic energy or kinetic energy final minus
kinetic energy initial or … one half mass times velocity final squared minus one half mass times
velocity initial squared. … We can factor out one half and mass. … And substitute in our values to
get one half times 5 times the quantity 2i plus 6j quantity squared minus the quantity
4i minus 3j quantity squared. … How do we square a unit vector expression?
That is a fair question. Let’s look at the example xi plus yj quantity squared to make
sure we understand this. Realize to square something means to multiply it by itself,
in the case of unit vectors, that means the dot product. So, … we can FOIL this
(First, Outer, Inner, Last) Right. and get x i dot x i plus x
i dot y j plus y j dot x i plus y i dot y i. … The dot product of x i and x i is just x squared
because the two unit vector i's are in the same direction. … The dot product of anything i and
anything j is just zero because the directions of i and j are perpendicular to one another. … And
the dot product of y j and y j is just y squared, again because unit vector j and j are in the same
direction. … So, this works out to be that we just square each of the unit vector coefficients,
however, it is important you realize why that is. … Alright, Billy, please finish the problem.
Absolutely. The net work then equals one half times 5 times the quantity
2 squared plus 6 squared minus the quantity 4 squared minus 3 squared.
That equals … well one half times 5 times the quantity … 4 + 36 or 40 minus the quantity
16 minus 9 or 7 and that equals 82.5 or … oh, 82 joules with 2 sig figs. I love
the arcane rounding rule! … This is not correct.
… Uh. That’s what I got with my calculator. Yeah, me too.
Well, it is still not correct. … I bet it is a negative.
It’s always a negative. … Oh, it’s the negative in
the negative 3 j in the initial velocity. Right. That needs to be negative 3 squared
because it’s the dot product of negative 3 j and negative 3 j. And squaring negative 3 gives us
positive 9. So, … it ends up being 40 minus 16 plus 9 or 25 instead of 40 minus 7 and we get
37.5 or 38 joules with 2 sig figs.
There we go. We always have to be careful with those negatives. … Lastly please, do not forget
the Net Work equals Change in Kinetic Energy Theorem. It may come in very handy. Oh and,
this equation is not on the AP Physics equation sheet. So, please memorize it because you will
likely need it. … Thank you very much for learning with me today, I enjoyed learning with you.