Physics Conceptual Questions - Mixed


by threewingedfury
Tags: conceptual, mixed, physics
threewingedfury
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#1
Dec9-06, 10:12 AM
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A block of mass m is pulled at constant velocity along a rough horizontal floor by an applied force T which is directed at an angle theta above the horizontal. The magnitude of the frictional force is:
a. T cos theta
b. T sin theta
c. 0
d. mu x mg
e. mu x mg cos theta

I figured it was d because frictional forces are mu x Force normal, but I didnt know because there is no mu in the problem.



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 theres no work done by the external force then I figured the energy would be conserved.
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AlephZero
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#2
Dec9-06, 10:24 AM
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d is wrong in the first one. Draw a picture of all the forces acting on the body. The reaction force between the block and the floor is not mg.

c is right for the second one if you are just considering mechanical forces and energy. Otherwise the answer is e. For example, a toy car driven by a solar powered electric motor. There's no "external force" on the system but, its KE + PE will change.
threewingedfury
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#3
Dec9-06, 10:27 AM
P: 29
so for 1 - e? because of the angle to the floor?

AlephZero
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#4
Dec9-06, 10:37 AM
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Physics Conceptual Questions - Mixed


No. Draw the picture. The reaction force is mg - T sin theta so the friction force would be mu(mg - T sin theta). But that's not one of the answers.
Ja4Coltrane
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#5
Dec9-06, 10:41 AM
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forgive me if I am not seeing something obvious, but I don't think any of those choices are right--it should be (mg-Tsin@)(mu)
threewingedfury
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#6
Dec9-06, 10:43 AM
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oh lord - well, Im just going by what I have on the question
threewingedfury
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#7
Dec9-06, 11:03 AM
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So which one out of the list would be close?
cristo
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#8
Dec9-06, 11:14 AM
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e is the answer. Consider drawing the diagram, letting theta be the angle of the plane to the horizontal, and alpha be the angle of the line of action of the weight of the block to the plane. Then [tex] \alpha=\frac{\pi}{2}-\theta [/tex]. Now, let beta be the top angle of the right angled triangle formed ny extending the line of action of the normal force to the horizontal below (i.e. the angle through which the weight must be resolved. Then [tex] \beta+\alpha=\frac{\pi}{2} \Rightarrow \beta+\frac{\pi}{2}-\theta=\frac{\pi}{2} \Rightarrow \beta=\theta [/tex]. Thus, the normal force is [tex]mg\cos\theta [/tex]. T is acting in the direction of the motion, and so does not feature in the normal force. Hence the frictional force is [tex] \mu mg\cos\theta [/tex], choice e
threewingedfury
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#9
Dec9-06, 11:36 AM
P: 29
thank you!
threewingedfury
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#10
Dec9-06, 11:59 AM
P: 29
ok 2 new questions if you guys dont mind

The center of mass of Earth's atmosphere is:
a. a little less than halfway between Earth's surface and the outer boundary of the atmosphere
b. near the surface of Earth
c. near the outer boundary of the atmosphere
d. near the center of the Earth
e. none of the above

The center of mass of a system of particles has a constant velocity if:
a. the forces exerted by the particles on each other sum to zero
b. the external froces acting on particles of the system sum to zero
c. the velocity of the center of mass is initially zero
d. the particles are distributed symmetrically around the center of mass
e. the center of mass is at the geometric center of the system

For the first one, I figured the a would be correct but it seemed too easy since its a concept question, so maybe the CoM is near the surface of Earth?

And for the second one, d - seemed most logical
cristo
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#11
Dec9-06, 12:33 PM
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A hint for the second question:
Suppose [tex]\bold{F}_{ij} [/tex] denotes the force exerted by the jth particle on the ith particle. Newton's 3rd law gives [tex]\bold{F}_{ij}+\bold{F}_{ji}=0. [/tex]. For the ith particle, Newtons 2nd law gives [tex] m_i\ddot{\bold{r}_i}=\sum_{j \neq i} \bold{F}_{ij}+ \bold{F}_i [/tex] where [tex] \bold{F}_i [/tex] is the external force on the particle. Do you know how to define the position vector of the centre of mass? Differentiating this yields the result. (n.b. the answer is not d)

For the first question, I'd go with your second thoughts- for example, what do you know about the atmosphere as we travel further away from the earth's surface?
AlephZero
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#12
Dec9-06, 01:00 PM
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Re the friction problem, Cristo answered the wrong question. The plane here is horizontal, not at an angle. He did give the right answer to the wrong question, though.

The right answer (IMHO): the horizontal component of the applied force is T cos theta. The block is moving at constant velocity so the resultant horizontal force is zero. Therefore, the friction force is also T cos theta (in the opposite direction).

The answer mu(mg-Tsin theta) is also right - except it doesn't appear in the list of choices.

Re the "CG of the earth atmosphere" question, the question is about the whole atmosphere, not just the bit of it which is above your own head. Think about a sphere with a thin layer of material (atmosphere) covering the whole surface.
cristo
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#13
Dec9-06, 01:10 PM
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Apologies.. I did read the question wrong! Thanks for pointing that out Alephzero. Both answers are correct! Sorry for any confusion caused!

The cg of the earths atmosphere threw me then as well- I didn't think about the whole atmosphere. Now you put it that way, the answer falls out!
vanesch
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#14
Dec9-06, 01:51 PM
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For the second question, (e) is the right answer. The reason is that the other options:
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
all have a problem.

(a) is wrong, because there can be external forces, as long as they are derived from a potential, and that potential is included in the potential energy (think: falling stone).
(b) is wrong: there's no need for a closed path.
(c) is wrong: the work done on a falling stone by the (external) gravity force is not zero, nevertheless PE + KE is conserved
(d) is wrong: when the external forces are NOT derivable from a potential (for instance, friction forces), then PE+KE is NOT conserved.
cristo
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#15
Dec9-06, 02:03 PM
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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.
threewingedfury
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#16
Dec9-06, 02:16 PM
P: 29
ok, now Im really confused, so many answers lol - was I right or wrong on the answers I suggested?
threewingedfury
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#17
Dec9-06, 04:42 PM
P: 29
ok so for the friction problem - T cos theta?

KE + PE problem - only when the work done by the resultant external force is zero?

Earths CoM - a little less than halfway between Earth's surface and the outer boundary of the atmosphere?

particles - the forces exerted by the particles on each other sum to zero?


Theres so many answers on here that I got confused
OlderDan
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#18
Dec9-06, 05:06 PM
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Quote Quote by cristo View Post
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.
That is not the correct interpretation of your equation. Want to restate it?


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