### Video Transcript

Two masses π and 88 grams are
attached to the ends of a light string passing over a smooth pulley. Determine the value of π, given
that when the system was released, the other mass descended 11.76 meters in two
seconds. Take the acceleration due to
gravity π equal to 9.8 meters per square second.

We begin by modeling the system as
shown. We are told that the masses are
equal to π and 88 grams. We recall that there are 1,000
grams in one kilogram, and as the gravity is given in meters per square second, we
need the mass to be in kilograms. 88 grams is equal to 0.088
kilograms. This means that the downward forces
of the two bodies are ππ and 0.088π. We have a light string and a smooth
pulley, which means that the vertical tensions will be equal. The entire system will move with
the same acceleration, and we are told that the 88-gram body descends. This means that the body of mass π
will accelerate upwards.

We can now use Newtonβs second law,
πΉ equals ππ, to create equations for body A and body B. Body A is accelerating
vertically upwards. Therefore, the sum of its forces is
equal to π minus ππ. This is equal to ππ. Body B is accelerating
downwards. Therefore, the sum of its forces is
0.088π minus π. We assume the downward direction is
positive. This is once again equal to the
mass multiplied by the acceleration. At this stage, we have three
unknowns: the tension, the mass of body A, and the acceleration of the system. We will need to use the information
from the question that tells us that the body of 88 grams descends 11.76 meters in
two seconds.

We can use the equations of uniform
acceleration to calculate the acceleration. These are often referred to as the
SUVAT equations, where π is the displacement β in this case equal to 11.76
meters. π’ is the initial velocity, π£ is
the final velocity, π is the acceleration, and π‘ is the time, in this question,
two seconds. We know that π is equal to π’π‘
plus a half ππ‘ squared. Substituting in our values, we have
11.76 is equal to zero multiplied by two plus a half multiplied by π multiplied by
two squared. The right-hand side simplifies to
two π, so this is equal to 11.76. Dividing both sides of this
equation by two gives us π is equal to 5.88. The acceleration of the system,
once it is released, is 5.88 meters per square second.

Substituting this as well as π
equals 9.8 into our equation for body A gives us π‘ minus 9.8π is equal to
5.88π. Adding 9.8π to both sides of this
equation gives us π‘ is equal to 15.68π. We will call this equation one. Substituting the value for the
acceleration into the equation for body B gives us 0.8624 minus π is equal to
0.51744. Rearranging this equation gives us
a value of π equal to 0.34496. We can now substitute the tension,
which is equal to 0.34496 newtons, into equation one. This tension force is equal to
15.68π. Dividing both sides by 15.68 gives
us π is equal to 0.022. This is the mass of the body in
kilograms. By multiplying this answer by
1,000, we can convert to grams, which is the same units as the other mass. The mass π is equal to 22
grams.