Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of
mass m1, in meters per square second.
1 answer:
Answer:
a1 = 3.68m/s²
Explanation:
Given values:
Mass of the block placed on the table, m1 = 12.25 kg
Mass of the block hanging vertically, m2 = 7.5 kg
Acceleration due to gravity, g = 9.8 m/s2
Tension in the string is T
Let the acceleration of mass 1 and mass 2 be a1 and a2
a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,
a = m2*g/(m1 + m2)
a = 7.5 x 9.8 / (12.5 + 7.5)
a = 3.68 m/s²
a1 = a2 = 3.68m/s²
a1 is directed to the right and a2 is directed downwards
Below is a diamonds to show the geometrical arrangements of both masses
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2) The mass of the raft is the product between its density, d, and its volume, V:
3) The weight of the raft is the product between its mass m and the gravitational acceleration,
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4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
where
is the water density and 
is the volume of displaced fluid.
The density of the raft (
) is smaller than the water density (
), this means that initially the buoyancy (which has upward direction) is larger than the weight (downward direction) and so the raft is pushed upward, until it reaches a condition of equilibrium and it floats. At equilibrium, the weight and the buoyancy are equal and opposite in sign:
and therefore, the apparent weight will be zero:
5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
When the raft is completely immersed in the water, the volume of fluid displaced is equal to the volume of the raft, 
. Therefore the buoyancy in this situation is
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
where
is the mass of the displaced water. From the previous equation, we obtain that 
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7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship
we can rewrite 
as:
and so
8) The volume of water displaced is (point 7)
. This volume is now "filled" with part of the volume of the raft, therefore is also the volume of the raft below the water level. We can calculate the fraction of raft's volume below water level, with respect to the total volume of the raft, 
:
Viceversa, the volume of raft above the water level is
. Therefore, the fraction of volume of the raft above water level is
9) Let's repeat steps 5-8 replacing
, the water density, with 
, the ethanol density.
9-5) The buoyant force is given by:
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.
9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
9-7) Using the relationship
, we can find the volume of displaced ethanol:
9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced:
. Therefore, the fraction of raft's volume below the ethanol level is
Consequently, the raft's volume above the ethanol level is
and the fraction of volume above the ethanol level is
2) The mass of the raft is the product between its density, d, and its volume, V:
3) The weight of the raft is the product between its mass m and the gravitational acceleration,
4) The apparent weight is equal to the difference between the weight of the raft and the buoyancy (the weight of the displaced fluid):
where
The density of the raft (
and therefore, the apparent weight will be zero:
5) The buoyant force B is the weight of the displaced fluid, as said in step 4):
When the raft is completely immersed in the water, the volume of fluid displaced
However, as we said in point 4), the raft is pushed upward until it reaches equilibrium and it floats. At equilibrium, the buoyancy will be equal to the weight of the raft (because the raft is in equilibrium), so:
6) At equilibrium, the mass of the displaced water is equal to the mass of the object. In fact, at equilibrium we have W=B, and this can be rewritten as
where
7) Since we know that the mass of displaced water is equal to the mass of the raft, using the relationship
and so
8) The volume of water displaced is (point 7)
Viceversa, the volume of raft above the water level is
9) Let's repeat steps 5-8 replacing
9-5) The buoyant force is given by:
when the raft is completely submerged. Then it goes upward until it reaches equilibrium and it floats: in this condition, B=W, so the buoyancy is equal to the weight of the raft.
9-6) Similarly as in point 6), the mass of the displaced ethanol is equal to the mass of the raft:
9-7) Using the relationship
9-8) The volume of raft below the ethanol level is equal to the volume of ethanol displaced:
Consequently, the raft's volume above the ethanol level is
and the fraction of volume above the ethanol level is