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1. Homework Statement
Assume that the block on the table (Figure 1) has twice the inertia of the hanging block.
(a)You give the block on the table a push to the right so that it starts to move. If the magnitude of the frictional force exerted by the table on the block is half the magnitude of the gravitational force exerted on the hanging block, what is the acceleration of the hanging block after you have stopped pushing the block on the table?
(b)What would this hangingblock acceleration be if you had pushed the block on the table to the left instead? Consider only a short time interval after you stop pushing.
2. Homework Equations
F_{net} = ma
F_{g} = mg
3. The Attempt at a Solution
I did get an answer to part(a). I solved as follows
F_{table} = T  .5mg = 2ma
F_{mass} = T  mg = ma
(set equations equal to T)
ma + mg = 2ma + .5mg
a + g = 2a + .5g
3a = .5g
a = 1/6g
And that was correct. I assumed the initial force to get the system moving had no part to play in the equations.
When I looked at part(b), I thought it would be the exact same, the forces in my equations had not changed even though there was a initial force in the x direction that was no longer present. However, the answer is not the same, and I don't understand how that momentary initial force impacts the final acceleration.
please advise,
arturo
Assume that the block on the table (Figure 1) has twice the inertia of the hanging block.
(a)You give the block on the table a push to the right so that it starts to move. If the magnitude of the frictional force exerted by the table on the block is half the magnitude of the gravitational force exerted on the hanging block, what is the acceleration of the hanging block after you have stopped pushing the block on the table?
(b)What would this hangingblock acceleration be if you had pushed the block on the table to the left instead? Consider only a short time interval after you stop pushing.
2. Homework Equations
F_{net} = ma
F_{g} = mg
3. The Attempt at a Solution
I did get an answer to part(a). I solved as follows
F_{table} = T  .5mg = 2ma
F_{mass} = T  mg = ma
(set equations equal to T)
ma + mg = 2ma + .5mg
a + g = 2a + .5g
3a = .5g
a = 1/6g
And that was correct. I assumed the initial force to get the system moving had no part to play in the equations.
When I looked at part(b), I thought it would be the exact same, the forces in my equations had not changed even though there was a initial force in the x direction that was no longer present. However, the answer is not the same, and I don't understand how that momentary initial force impacts the final acceleration.
please advise,
arturo
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