Physics Opted

[despaill]

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Hello everyone,
I was wondering if anyone could explain questions 11 and 39 in the physics section of the opted sample test?
I feel like they're simple, but can't seem to get the correct answer :-(

11. A small girl applies a horizontal force of 2 N to a 10 N box which slides across the floor with a constant speed of 3 m/s. What is the frictional force, in N, by the floor on the box?

39. What is the focal length in cm of a lens that produces an image 30 cm behind it when the object is placed 6 cm in front of it?

I did not make up these questions, they are from the opted test.

A quick response would be extremely appreciated!!!
Thanks in advance!!

 

[achirum]

Hello everyone,
I was wondering if anyone could explain questions 11 and 39 in the physics section of the opted sample test?
I feel like they're simple, but can't seem to get the correct answer :-(

11. A small girl applies a horizontal force of 2 N to a 10 N box which slides across the floor with a constant speed of 3 m/s. What is the frictional force, in N, by the floor on the box?

39. What is the focal length in cm of a lens that produces an image 30 cm behind it when the object is placed 6 cm in front of it?

I did not make up these questions, they are from the opted test.

A quick response would be extremely appreciated!!!
Thanks in advance!!

Is that the whole question for #11? No coefficient of kinetic friction?

 

[despaill]

Yup, that's it..

 

[Dena1342]

Hello everyone,
I was wondering if anyone could explain questions 11 and 39 in the physics section of the opted sample test?
I feel like they're simple, but can't seem to get the correct answer :-(

11. A small girl applies a horizontal force of 2 N to a 10 N box which slides across the floor with a constant speed of 3 m/s. What is the frictional force, in N, by the floor on the box?

39. What is the focal length in cm of a lens that produces an image 30 cm behind it when the object is placed 6 cm in front of it?

I did not make up these questions, they are from the opted test.

A quick response would be extremely appreciated!!!
Thanks in advance!!

For 39:

1/O + 1/i= 1/f
1/6 + 1/30 = 1/f sooo...
6/30 = 1/f...which means
1/5 = 1/f, f= 5

 

[Slapshot13]

#11. Constant speed means a=0. therefore fnet= 0..thus if a 2N Force is applied to the box, for Fnet to be 0 a force of equal magnitude and opposite direction (friction) must exist. Thus, The answer should be 2. The weight of the box means absolutely nothing. this question covers both newtons first and second law.

#39. using the formula
1/i +1/o = 1/f where i= image length and o= object length.. (Solve for f= focal length)


Then again, I may be wrong..But Im confident these are the answers..

 

[blysssful]

11. A small girl applies a horizontal force of 2 N to a 10 N box which slides across the floor with a constant speed of 3 m/s. What is the frictional force, in N, by the floor on the box?

39. What is the focal length in cm of a lens that produces an image 30 cm behind it when the object is placed 6 cm in front of it?

I haven't taken the opted test yet, so I haven't looked at the answer choices... but here's what I think:
11. constant speed= 0 acceleration = 0 Net Force... therefore if there are only two forces acting on the box in the x direction (the push and kinetic friction), they should be equal. So, the frictional force is equal to 2N in the direction opposite from the initial force.

39. (1/focal length) = (1/image distance) + (1/object distance)
(1/f) = (1/30) + (1/6) = (1/30) + (5/30)
(1/f) = (6/30)
f = 5cm

Sorry if those aren't right. It's how I'd work them out, anyway.

Edit: Oops, I guess I took a while to respond!

 

[achirum]

Yeah, the answers given for 11 sound absolutely correct here.

Edit:

Just kind of pondering things however, since the box is moving, that means kinetic friction is at work correct? And if it is kinetic friction that is at work, then the total force cannot possible = 0 (I know according to Netowns Law since a=0 then F must = 0 but I'm just saying) as that would mean the box isn't moving. In this case it would be static friction at work.

Am I thinking about this incorrectly?

 

[Slapshot13]

Yeah, the answers given for 11 sound absolutely correct here.

Edit:

Just kind of pondering things however, since the box is moving, that means kinetic friction is at work correct? And if it is kinetic friction that is at work, then the total force cannot possible = 0 (I know according to Netowns Law since a=0 then F must = 0 but I'm just saying) as that would mean the box isn't moving. In this case it would be static friction at work.

Am I thinking about this incorrectly?

The question is stated such that the box is already in motion, and therefore we can disregard how it came to getting into motion (like you said, having a net force not= 0). When Fnet= 0, it does not mean that the object is not moving, it also means the objects motion is constant (there is no change or acceleration). It is then, very plausible to have motion when Fnet=0 (Assuming some net force was used originally to propel the object into motion).

That's a mouthful, I think I'm just telling my self what I want to hear- so ignore that haha.

 

[blysssful]

The question is stated such that the box is already in motion, and therefore we can disregard how it came to getting into motion (like you said, having a net force not= 0). When Fnet= 0, it does not mean that the object is not moving, it also means the objects motion is constant (there is no change or acceleration). It is then, very plausible to have motion when Fnet=0 (Assuming some net force was used originally to propel the object into motion).

That's a mouthful, I think I'm just telling my self what I want to hear- so ignore that haha.

Hahaha physics is always a mouthful. But you're definitely right.

"An object in motion will stay in motion unless there is a net force acting on it" (or however it goes 😛)

The question isn't asking about how much force the girl needed to give to initially get the box to move (i.e. how much force to overcome static friction). It's asking how much force counters her force if there is no acceleration. No acceleration = 0 net force. No exceptions (well there may be exceptions, but not for problems like this).