Physics Conceptual Questions - Mixed
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cristo said:I disagree... Consider the postion vector of the COM. [tex]M \bold{R} =\sum_i m_i \bold{r}_i \[/tex] Differentiating twice yields [tex]M \bold{\ddot{R}} =\sum_i m_i \ddot{r_i} = \sum_i\sum_{j\neq i} \bold{F}_{ij} + \sum_i \bold{F}_i[/tex] Now, since [tex]\bold{F}_{ij}+\bold{F}_{ji}=0.[/tex] the first term on the rhs vanishes, hence [tex]M \bold{\ddot{R}} = \sum_i\bold{F}_i[/tex] So, in order for the COM to have constant velocity, the sum of all external forces acting on the ith particle must be zero.
You should read more carefully. I was talking about the SECOND question, and nobody is talking about a uniform motion of the COG, but rather about the conservation of KE + PE.
This question:
The sum of the kinetic and potential energies of a system of objects is conserved:
a. only when no external force acts on the objects
b. only when the objects move along closed paths
c. only when the work done by the resultant external force is zero
d. always
e. none of the above
Is c correct? If there's no work done by the external force then I figured the energy would be conserved.
Here, I maintain that the correct answer is e, for the reasons I mentionned before.
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threewingedfury said:So the T cos theta isn't right?
It IS the right answer. In fact, given that:
1) the motion of the block is uniform
2) the forces acting on it are
- weight (vertical)
- binding force of the surface (vertical)
- friction force (horizontal)
- known force T under angle T
we must have that the total force is 0 (because uniform motion), and hence that the horizontal and vertical components are 0.
The vertical component is given by the sum of the weight, the vertical component of T and the binding force, but we don't care.
The horizontal component is given by the sum of the friction force and the horizontal component of T, which equals T cos theta.
So this means that the friction force must equal T cos theta (in the other direction).
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vanesch said:You should read more carefully. I was talking about the SECOND question, and nobody is talking about a uniform motion of the COG, but rather about the conservation of KE + PE.
This question:
Here, I maintain that the correct answer is e, for the reasons I mentionned before.
I probably should not be speaking for cristo, but I don't think he was disagreeing with you. I think he was responding to threewingedfury, but since he did not quote the post to which he was replying, the sequence suggested otherwise.
So to all who read here, PLEASE avoid creating such a jumpled thread by keeping problems in separate threads, and please identify the posts to which you are responding if it is not the one directly above yours.
Ja4Coltrane
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hey christo, sorry to go back to something old but on that first problem, it seems that choice E is only right if T=mg. this bothers me.
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OlderDan said:I probably should not be speaking for cristo, but I don't think he was disagreeing with you. I think he was responding to threewingedfury, but since he did not quote the post to which he was replying, the sequence suggested otherwise.
So to all who read here, PLEASE avoid creating such a jumpled thread by keeping problems in separate threads, and please identify the posts to which you are responding if it is not the one directly above yours.
Yes, olderdan is correct, apologies for the way it appeared!
Ja4Coltrane said:hey christo, sorry to go back to something old but on that first problem, it seems that choice E is only right if T=mg. this bothers me.
As pointed out above, I answered the wrong question. Tcos theta is the answer.
Now I've worked out how to quote more than one thing, I will do in future!
edward123
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i think it should be T cos [tex]\theta[/tex]
bcoz the body is moving with constant velocity and so acc to the Newton's first law of motion it is moving under the balanced forces
bcoz the body is moving with constant velocity and so acc to the Newton's first law of motion it is moving under the balanced forces
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