Video Transcript
A body of mass 200 grams rests on a
rough horizontal table. It is connected by a light
inextensible string passing over a smooth pulley fixed to the edge of a table to
another body of the same mass hanging freely below the pulley two centimeters above
the ground. The coefficient of friction between
the table and the body resting on it is one-third. Given that the system was released
from rest and the hanging body descended until it hit the ground, how much further
did the body on the table travel until it came to rest? Take acceleration due to gravity 𝑔
to be equal to 9.8 meters per square second.
There is a lot of information in
this question, so we will begin by drawing a diagram. We have two bodies of the same mass
of 200 grams. The weight of these bodies is equal
to the mass multiplied by gravity. As there are 1000 grams in one
kilogram, these both have a downward force of 0.2𝑔. The tension throughout the string
will be equal. And when the system is released,
the freely hanging body will accelerate downwards and the body on the table will
accelerate to the right. As the horizontal table is rough,
there will be a frictional force acting against the body. There is also a reaction force
going vertically upwards.
We know that the frictional force
is equal to 𝜇 multiplied by the reaction force when 𝜇 is the coefficient of
friction, in this case, one-third. Newton’s second law states that
force is equal to mass multiplied by acceleration. We can use this to resolve
vertically and horizontally. The body on the table is not moving
vertically. Therefore, the forces up must equal
the forces down. We have 𝑅 is equal to 0.2𝑔. The friction force must be equal to
a third of this. This gives us a friction force of
49 over 75 newtons. Adding in the fact that the freely
hanging body is two centimeters from the ground, we will now clear some space for
the remainder of our calculations.
Resolving horizontally for the body
on the table, we see that the sum of the forces is equal to 𝑇 minus 𝐹 r. This is equal to 0.2𝑎. We can substitute in the value of
friction here. The freely hanging particle is
accelerating downwards. This gives us the equation 0.2𝑔
minus 𝑇 is equal to 0.2𝑎. We have a pair of simultaneous
equations that we can solve to calculate the acceleration 𝑎. Adding equation one and equation
two gives us 98 over 75 is equal to 0.4𝑎. Dividing both sides by 0.4 gives us
an acceleration of 49 over 15. Once released, the system moves
with an acceleration of 49 over 15 meters per second squared.
After the freely hanging body hits
the ground, the string becomes slack and its tension is equal to zero. We can then calculate the
acceleration of the body on the table after this point by solving negative 49 over
75 is equal to 0.2𝑎. Dividing both sides of this
equation by 0.2 gives us 𝑎 is equal to negative 49 over 15. We notice that the acceleration
after the string becomes slack is the negative acceleration prior to it.
Let’s now consider what is
happening to the body on the table throughout its motion. We know that it starts from rest
and accelerates uniformly at 49 over 15 meters per square second. During this time, it travels a
distance of two centimeters as it must travel the same distance as body B. After the freely hanging body hits
the ground, the body on the table decelerates. Its acceleration is equal to
negative 49 over 15 meters per square second. We need to calculate how far it
travels before it comes to rest. The final speed of the first part
of the journey is equal to the initial speed of the second part.
We can now use our equations of
uniform acceleration or SUVAT equations to calculate 𝑠. We know the following values for
the two parts of the journey. We can now substitute these values
into the equation 𝑣 squared is equal to 𝑢 squared plus two 𝑎𝑠. In the first part, as 𝑢 is equal
to zero, we have 𝑣 squared is equal to two multiplied by 49 over 15 multiplied by
two. In the second part, as 𝑣 is equal
to zero, we have zero is equal to 𝑣 squared plus two multiplied by negative 49 over
15 multiplied by 𝑠.
We can solve these simultaneously
to eliminate 𝑣 squared. We can then divide our new equation
by two and 49 over 15. Two minus 𝑠 is equal to zero,
which means that 𝑠 is equal to two. The body travels another two
centimeters until it comes to rest. This is the same distance as in the
first part of the journey as the acceleration and then deceleration are equal.
