insight quick question

[LinusX]

I did the question (attached) but I am a little confused about it. Shouldnt we include the mass of the hanging mass in our acceleration calulations? I got the net force (8000-friction-force on container) but then shouldnt we divide the force by (hanging mass + container)?

Thanks in advance,
LinusX

Oh yeah, the solutions dont do this

[golden]

Wait a minute how did the solutions even get this?
Question 3
If the hanging mass, mH
Worked solution , is increased to 1000 kg and the friction force stays the same, what is the magnitude of the container’s acceleration now?

ΣF = ma
10 000 – (1000 + 7000) = 1400a
Therefore, a = 1.43 m s–2.

I don't get why they took off the 1000 at all.

[LinusX]

Yeah I thought it was a bit like the pulley question, and the net force needs to act to pull both the masses....Anyone care to explain?

[golden]

Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?

[Lasercookie]

Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?

I got C as my answer as well.

The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.

[r_damico]

i think it might be because rather the friction acting up the slope, it is down the incline as it is opposing the motion of the container

ΣF = F_driving - F_retarding

where F_retarding = weight component + friction

[golden]

Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?

I got C as my answer as well.

The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.

And I suppose question 8 is wrong too? I got 2.98 x 10^-14 J.

Could someone please explain Q12.

Two ships go at 0.65 c away from each each other.
What is the speed of one ship as measured by the other?

[LinusX]

Can anyone explain this question? I still dont understand why you dont use both the masses in the calculation as the net force has to act on both the objects to give a shared acceleration.

[thatricksta]

Wait a minute how did the solutions even get this?
Question 3
If the hanging mass, mH
Worked solution , is increased to 1000 kg and the friction force stays the same, what is the magnitude of the container’s acceleration now?

ΣF = ma
10 000 – (1000 + 7000) = 1400a
Therefore, a = 1.43 m s–2.

I don't get why they took off the 1000 at all.

this is the answer I got
the previous question indicated that when fnet = 0 and m_h is the only force acting upon the object, the friction force is 8000N. when increasing the applied force on the object to 10,000 N, the frictional force is still the same (8000N)

Question 7 from Relativity on Insight 2010:
Shouldn't it be C not D?

I got C as my answer as well.

The worked solutions for Insight 2010 also show the answer for C...yet they wrote D in the box.

And I suppose question 8 is wrong too? I got 2.98 x 10^-14 J.

Could someone please explain Q12.

Two ships go at 0.65 c away from each each other.
What is the speed of one ship as measured by the other?

This one is not as complicated as it seems.

the formula for this is


So, lets say Freddie is going at w and Khokho is going at velocity v
because velocity is a vector, in respect to Freddie, Khokho is going at a velocity of -v

therefore you will get

= 0.91c

I also got C for question 8.

[Vincezor]

Can anyone explain this question? I still dont understand why you dont use both the masses in the calculation as the net force has to act on both the objects to give a shared acceleration.

Yeah I'm wondering as well...

I get

As F = ma

?

[thatricksta]

Can anyone explain this question? I still dont understand why you dont use both the masses in the calculation as the net force has to act on both the objects to give a shared acceleration.

Yeah I'm wondering as well...

I get

As F = ma

?

In relation to tension, components are isolated.
I don't believe this question was designed to act as a pulley, this is just regular motion with a net force similar to an engine if you want to think of it like that.

[Vincezor]

In relation to tension, components are isolated

Sorry I still don't get it. Why aren't both masses accelerating at the same rate? The question doesn't ask for tension? :S

[thatricksta]

In relation to tension, components are isolated

Sorry I still don't get it. Why aren't both masses accelerating at the same rate? The question doesn't ask for tension? :S

Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.

[Vincezor]

Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.

Oh I think I get it now. So does this mean the hanging block is moving at a constant speed down? (a=0)?

[thatricksta]

Just take the falling mass as the driving force acting on the object... I do this every time, the net force comes from the tension of the string caused by the falling object.

Oh I think I get it now. So does this mean the hanging block is moving at a constant speed down? (a=0)?

a = gravity, so F=ma

I tend not to over think it too much, rather just look at it and be like OK, gravity, mass, F=ma.