### Video Transcript

An object is acted on by three
simultaneous forces. π
one equals negative 3.00π’ plus
2.00π£ newtons, π
two equals 6.00π’ minus 4.00π£ newtons, and π
three equals
2.00π’ plus 5.00π£ newtons. The object experiences acceleration
of 4.23 meters per second squared. The object is initially at
rest. Find the mass of the object. Find the speed of the object after
5.00 seconds. Find the components of the velocity
of the object after 5.00 seconds.

Given the three forces π
one, π
two, and π
three acting on an object initially at rest, which result in an
acceleration of 4.23 meters per second squared, we want to solve first for the mass
of the object. We can label that π. We also want to solve for the speed
the object attains after 5.00 seconds. Weβll call this speed π . And finally, we want to solve for
the velocity components of the object after that same amount of time. Weβll call these components π£ sub
π₯ and π£ sub π¦.

To start on our solution, we can
draw a sketch of the forces acting on this object. If we sketch in the three forces
acting on our object on a pair of axes in units of newtons, we see that we can add
these together algebraically to solve for the net force acting on the object. Writing out πΉ one, πΉ two, and πΉ
three by their component parts, when we add them together, we find a net resultant
force of 5.00π plus 3.00π newtons.

Now that we know the net force, we
can use Newtonβs second law, which says this net force equals the objectβs mass
times its acceleration to solve for object mass π. By the second law, object mass is
equal to the magnitude of the net force acting on it divided by its acceleration
π. We solve for the magnitude of the
net force by squaring each of its two components, adding the results together, and
taking the square root of that sum.

Dividing that magnitude by our
acceleration, given as 4.23 meters per second squared, we find that object mass, to
three significant figures, is 1.38 kilograms. Thatβs the mass of the object under
the influence of these three forces.

Next, we want to solve for the
speed of our object after a time of 5.00 seconds has elapsed. Since the acceleration, in this
case, is a constant value across that time, this tells us that the kinematic
equations of motion apply to this objectβs motion. We look through these different
equations seeking a match between what weβre trying to solve for, final speed, and
the information we already know.

Since weβre told that time, which
we can call π‘, is 5.00 seconds, the first equation written down is a match for what
we want to solve for. Written in terms of our variables,
this equation is π is equal to the initial speed of the object plus its
acceleration times time. We know that because it started at
rest, its initial speed is zero. So, its final speed is simply π
times π‘, or 4.23 meters per second squared times 5.00 seconds. This is equal to 21.2 meters per
second. Thatβs the objectβs final speed
after 5.00 seconds under this acceleration.

Lastly, we want to solve for the
components of the velocity of the object at the same time π‘ equals 5.00
seconds. To start on that solution, letβs
draw the net force acting on our object in on our graph. Our net force, the sum of πΉ one,
πΉ two, and πΉ three, is pointed at some angle, we can call it π, from the positive
π₯-axis. This angle is important because the
direction of our net force by Newtonβs second law is the same as the direction of
the resulting acceleration. And since our particle started from
rest, that means its velocity will also be in the direction of force since itβs in
the direction of acceleration.

Hereβs what we know so far about
our final velocity at π‘ equals 5.00 seconds. We know, first of all, that its
magnitude is equal to π , which weβve solved for. And second, we know that this
velocity has an π₯- and a π¦-component. But we just donβt know what those
components are. However, if we solve for the
direction of the net force, then we can figure out how the final speed magnitude
divides up over the two components of velocity.

So, looking at our diagram, if we
draw in the π₯- and π¦-components of the net force and recognize we know their
numerical values from solving for them earlier, we can write that the tangent of our
angle π is equal to 3.00 newtons divided by 5.00 newtons. Taking the arctangent of both
sides, we now have an expression for the angle π. And when we solve for it, we find
itβs roughly equal to 31 degrees.

Now, that we know the angle π, we
can solve for the components of velocity based on the final magnitude of that
velocity, which is speed. We can write π£ sub π₯ as the
magnitude of the velocity, which is π , times the cos of π and π£ sub π¦ as π
times the sin of π. Plugging in for these values when
we solve for π£ sub π₯ and π£ sub π¦, we find theyβre equal, to three significant
figures, to 18.1 meters per second and 10.9 meters per second, respectively. In component form then, our final
velocity is 18.1π plus 10.9π meters per second.