HELP-attack of the physics question!!!!

[Sarahisme]

Hey i am currently studying for a physics exam and the questions we have been given have answers to only the odd numbered questions. But i like to be throuhgly prepared for exams and so am doing all of them. As you have probably guessed by now i need some help with an even numbered question. I think i have done it, but i am pretty unsure, so i would really really like to check if my answer is correct. I will attach the problem as a picture.

 

 

 

Thanks

Sarah

 

 

p.s

i wasn't sure whether to post here or in the physics:classical mechanics section of the forums. tell me if i should repost or whatever

[Sarahisme]

lol anyone? even a comment? like "this is too hard" or "man sarah you are stupid, this question is so easy!" anything at all

[swansont]

Two comments: 1. I don't see your answers, so there's nothing to check. 2. Patience. Two hours isn't really a long enough time to expect an answer.

[Sarahisme]

oh ok an answer sorry i got

 

a)-40N

b)720N in positive x direction

c)7.2m/s/s

 

sorry bout that

 

Sarah

[swansont]

oh ok an answer sorry i got

 

a)-40N

b)720N in positive x direction

c)7.2m/s/s

 

sorry bout that

 

Sarah

 

The small block is accelerating in the forward direction. What force is causing that acceleration?

 

The answer to a affects b. How did you come up with that answer? And you were supposed to give both the applied and net forces.

 

Your answer to c is consistent with b if that was the net force, but since b is incorrect, so is c.

[Sarahisme]

i think the fs (static friction) is causing the small lock to acclerate in the forward direction.

 

how about if i upload a picture(s) of my working?

[Sarahisme]

oh btw i just checked out some of your cartoons...lol quantum mechanic hahahaha

[Primarygun]

a)-40N

b)720N in positive x direction

Is b correct?

There should be a reaction force of the friction which act towards left on the 100kg.

F-f=net force.

[Sarahisme]

ok I have completely rethought the problem, ignore my previous answers....ok here goes I’ll try and explain my new answers....

 

a)

what causes the small block to move is the friction between it and the larger block. since the little block is a=4m/s/s, therefore the force required to move it must be F=ma (where m is mass of the little block = 20kg), and so the force required to accelerate an object is 80N. Since the acceleration of the small block is caused by the friction between it and the large one, therefore the frictional force exerted by the 100-kg mass on the 20-kg mass is 80N.

 

b)

The net force in the y direction is 0.

In the x direction the net force is the force required to accelerate a 100kg mass at 6m/s/s, but by Newton's third law there is a retarding force of 80N (opposite to the frictional force), so the net force in the x direction is

F - 80 = 100kg x 6m/s/s

which is equal to 680N

 

c)

This is I think the easy part.

With the same force of 680N what is the acceleration of the 100kg block? well...

F=ma

680N = 100kg x a

therefore a = 6.8m/s/s

So the acceleration after the 20kg mass falls off (assuming that the force F does not change) is 6.8m/s/s

[Sarahisme]

so what do people think? compltely wrong or what?

[swansont]

The applied force is 680 N, while the net force is 600 N. Other than that, I agree.

[Sarahisme]

i am not entirely sure what you mean, sorry :S which part of my answer?

[Sarahisme]

oh i see... sorry now that i think about it, so to summarise

 

a) is correct

but

b) should be 600N, because thats the net force

and c) should use 680N because thats the force applied to move the system in b) ?

[Sarahisme]

is that right? lol i really hope so this time

[swansont]

deleted - I misread the diagram. Incorrect analysis.

[The Rebel]

I still think the answer to a) is -40N

 

The acceleration between the two surfaces is 2ms-2. So the answer to a) is 20kg x -2ms-2 = -40N

 

I agree to the workings of b) and c) however. Only my answers would be

 

b) Net force = 720N, Force F = 760N

 

c) a = 7.6ms-2

[swansont]

I still think the answer to a) is -40N

 

The acceleration between the two surfaces is 2ms-2. So the answer to a) is 20kg x -2ms-2 = -40N

 

The block is accelerating at 4 m/s², and the only force on it in the x direction is friction.

 

It's not the relative acceleration that matters. If the relative acceleration were smaller, it would have to be because the frictional force was larger. If the relative acceleration was zero, it's not because the friction is zero. If the friction was zero, the top block would not slide at all, which would maximize the relative acceleration.

[swansont]

oh i see... sorry now that i think about it' date=' so to summarise

 

a) is correct

but

b) should be 600N, because thats the net force

and c) should use 680N because thats the force applied to move the system in b) ?[/quote']

 

Yes, I think this is correct.

[Sarahisme]

deleted - I misread the diagram. Incorrect analysis.

 

whats this about?

 

 

Oh thanks a million for helping me swansont

[swansont]

whats this about?

 

 

Oh thanks a million for helping me swansont

 

I goofed, and didn't want to leave a wrong answer around to confuse things.

[Sarahisme]

oh okies well still thanks for everything