Question

Physics

Answer 1

For a)
I said that the student's claim is correct because
this is because the movement of the block is determined by the person's push and the drag force on the block. So: person's push - drag = m*a
We see that the person's push ratio to the speed of the block is linear. This means that the air resistance has to be proportional to the speed as well as the force of the person's push., so that it satisfies newton's second law and if the force of air resistance was not proportional, then the speed wouldn't double from the push of 20N to 40N and from 40N to 80N, Newton's second law would be:
push-drag = m*a,
Acceleration is proportional to the net force and drag is part of the net force so the student's claim is correct

Answer 2

For b i) I drew an FBD with a force pointing to the right (Fa) and 2 forces pointing to the left (Fbp) which is force of box on person) and the friction force.
Like th

Answer 3

Now for part ii of B I am sort of confused on how to find the acceleration of the block after the person is released, I know that it is to the left but kind of confused on how to find the value.

Answer 4

@Hero @Shadow @dude @jhonyy9 @imqwerty

Answer 5

@vocaloid

Answer 6

When the person is in contact with the block, the block is moving at a constant speed of 5m/s. From the table, we can infer that the person is pushing the block with a constant force of 80N.

Answer 7

He is pushing on the block because there are retarding force (the drag) which try to slow down the block, so to maintain a constant velocity he has to push the block with an equal and opposite force ( 80N )

Answer 8

Since the person has to push with an equal and opposite force of 80N to maintain a speed of 5m/s, we can say that the drag would also be 80N at the moment when the speed is 5m/s

Answer 9

The person stops pushing and the drag at the moment is 80N. We can say that the net force is 80N in the direction opposite to the direction of motion. So the acceleration would be \(80/10 = -8m/s^2\) (F=ma)

Answer 10

Okay, thanks!
How would my other answers for this question be? correct or...

Answer 11

What is \(F_{FP}\) in the diagram? Force of friction on the person?

Answer 12

yes

Answer 13

I think I made a mistake though. Constant velocity so acceleration is zero. rip

Answer 14

And what is \(F_{AP}\)?

Answer 15

Applied force, I just labeled it that. Basically the force that gets the person moving.


If I did that correct

Answer 16

It's the friction that gets the person moving

Answer 17

Oh no. I am a big brain. The friction is what gets ppl moving and we learned that in class and I forgot. Actually a genius. I don't fully remember but would the friction would be towards the right canceling the force that the block puts on the person?

Answer 18

Wait a minute, shouldn't friction still be pointing to the left? Because for instance if I pulled an object that is on a table to the right then wouldn't friction act to the left?

Answer 19

The friction would act in the left direction. By "Applied force" do you mean the force applied by the ground/ice on the person?

Answer 20

yeah, thats what i mean

Answer 21

the force of ice on the person would have an x component and a y component. You've labeled the x component as \(F_{AP}\) and the y component as \(F_{NP}\) which works ig
I think the diagram is correct

Answer 22

wonder why I got 0 out of 2 for that one

Answer 23

it is all prolly cuz I said acceleration points to the right, this is the end of me

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight
wonder why I got 0 out of 2 for that one
\(\color{#0cbb34}{\text{End of Quote}}\)
It could be because of labeling? And yes, the acceleration that you've mentioned is def, not correct

Answer 25

Its the acceleration, and also I have another question that I'll post but I gtg so if u could just look at that anytime that would be great.

Answer 26

sure, tag me whenever you post it