### Video Transcript

Two blocks are connected across a
pulley by a rope as shown. The mass π one of the block on the
table is 4.0 kilograms. And the mass π two of the hanging
block is 1.0 kilogram. The mass of the rope is
negligible. The table and the pulley are both
frictionless. Find the acceleration of the
system. Find the tension in the rope. Find the speed of the hanging block
when it hits the floor. Assume that the hanging block is
initially at rest and located 1.0 meters vertically above the floor.

From this last bit of information,
we can add some detail to our diagram. We now know that π two is a
distance we can call π of 1.0 meters above the floor. Weβre told previously the masses π
one and π two. And we want to solve overall for
three pieces of information. We want to solve for the
acceleration π of the system, the tension π in the cable connecting the two
blocks, and the final speed, we can call π£ sub π of π two, as it hits the
ground. As we get started solving for
system acceleration π, we recall Newtonβs second law of motion that the net force
acting on an object or a system is equal to the mass of the object multiplied by its
acceleration.

When we consider the forces
involved with respect to our diagram, we see that the only force responsible for the
motion of this system of masses is the gravitational force acting on π two. Thatβs because our system overall
is frictionless and the gravitational force on π one is counteracted by the normal
force pushing up on π one. So by Newtonβs second law, we can
write that the weight force on π two, which is π two times π, equals the mass of
our system, π one plus π two, multiplied by the acceleration of the system,
π. Rearranging this expression to
solve for π, we see itβs equal to π two times π divided by the sum of the masses
π one and π two. π weβll treat as exactly 9.8
meters per second squared. And since weβre given π one and π
two, weβre ready to plug in and solve for π.

When we enter this expression on
our calculator, we find that π is 2.0 meters per second squared. Thatβs the overall acceleration of
this system of masses. Next, we want to solve for the
tension π in the line connecting the two masses. The interesting thing about solving
for π is that since π, the tension, is a force on both masses, we could look at
either one of the masses to solve for it. Just to pick a mass, letβs choose
π two. The two forces acting on π two are
the tension force π and the weight force π two times π. If we consider motion down to be
motion in the positive direction, then Newtonβs second law tells us we can write π
two times π minus π is equal to π two times π, where π is the acceleration we
solved for earlier.

Rearranging to solve for π, we
find itβs equal to π two times π minus π, or 1.0 kilograms multiplied by 9.8
minus 2.0 meters per second squared. When we multiply these values
together, we find that π is 7.8 newtons. Thatβs the tension in the line
connecting the masses. And finally, knowing that π two
starts from rest and descends a distance 1.0 meters before it reaches the floor, we
want to solve for π£ sub π, its final speed, just as it does reach the floor. Since the mass π two has a
constant acceleration, we can use the kinematic equations to solve for its
motion.

Specifically, we can use the
equation that π£ sub π squared equals π£ sub zero squared plus two times π times
π. Since π two starts at rest, that
means π£ sub zero is zero. So our equation is simplified to π£
sub π squared equals two times π times π. When we take the square root of
both sides and plug in 2.0 meters per second squared for π and 1.0 meters for π,
when we calculate the square root, to two significant figures, it equals 2.0 meters
per second. Thatβs the speed of block π two
just as it hits the ground.