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Question 16
Two masses, 2m kg and 4m kg, are attached by a light string. The string is placed over a smooth pulley as shown. The two masses are at rest before being released an... show full transcript
Step 1
Answer
Let ( g ) be the acceleration due to gravity acting on the masses.
For the 4m mass, the forces acting on it are:
Using Newton's second law, we have:
[ 4mg - T - kv_{4m} = 4m \cdot a ]
where ( a ) is the acceleration of the system.
For the 2m mass:
Thus:
[ T - 2mg - kv_{2m} = 2m \cdot (-a) ]
(because the acceleration acts in the opposite direction)
Setting the equations for acceleration equal based on the relationship ( a = \frac{dv}{dt} ), we can derive:
[ a = \frac{4mg - T - kv_{4m}}{4m} ]
[ a = \frac{T - 2mg - kv_{2m}}{-2m} ]
By solving these equations simultaneously, we arrive at:
[ \frac{dv}{dt} = \frac{gm - kv}{3m} ]
Step 2
Answer
Starting with the differential equation derived earlier:
[ \frac{dv}{dt} = \frac{gm - kv}{3m} ]
Rearranging gives:
[ \frac{dv}{gm - kv} = \frac{1}{3m} dt ]
Integrating both sides, we have:
[ \int \frac{1}{gm - kv} dv = \int \frac{1}{3m} dt ]
This yields:
[ -\frac{1}{k} \ln |gm - kv| = \frac{t}{3m} + C ]
Solving for ( v ) requires determining the constant ( C ) when the mass is initially at rest (( v = 0 ) at ( t = 0 )).
Substituting values, we solve for ( v ) at the specified time:
When ( t = \frac{3m}{k} \ln 2 ):
[ v = \frac{gm}{2k} ]
Thus, we have shown the required relationship.
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