HELP WITH AP PHYSICS- 5 min drill Multiple choice easy?

[girlphysics]

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated.


_____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

______2. A box of mass m slides on a horizontal surface with initial speed v0. It feels no forces other than it weight and the force from the surface. If the coefficient of kinetic friction between the box and surface is μ, how far does the box slide before coming to rest?
A) v02/2μg B) v02/μg C) 2v02/μg
D) mv02/μg E) 2 mv02/μg

_____3. An object initially at rest experiences a time-varying acceleration given by a = ( 2 m/s3 t) for t ≥ 0. How far does the oject travel in the first 3 seconds?
A) 9 m B) 12 m C) 18 m D) 24 m E) 27 m

______4. An engine provides 10kW of power to lift a heavy load at constant velocity a distance of 20 meters in 5 seconds. What is the mass of the load?
A) 100 kg B) 150 kg C) 200 kg D) 250 kg E)300 kg

______5. A ball (mass= .08 kg) is dropped onto the floor from a height of 2 m, and after bouncing off the floor rises almost to its original height. If the contact time with the floor was .04 s, what average force did the floor exert on the ball?
A) 0.16 N B) 16 N C) 25 N D) 36 N E) 64 N

 

[girlphysics]

For #4, I said Work/Time=Power so Fd/t=P so Ma*d/t=P because W=fd=ma*d so then I plugged in all the numbers and got .25 which leads me to believe its 250. ? I don't know I picked D

 

[cepheid]

Welcome to PF girlphysics!

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated. _____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

For this first problem, you can just use the relevant kinematics equation. What you want is the equation that tells you how distance varies with time for motion with constant acceleration. Start with that. Do you see any equations from your notes or your book that give distance vs. time for the case of a constant acceleration?

 

[girlphysics]

v0t+1/2at^2 so because initial is 0, just 1/2at^2??

 

[genericusrnme]

1)
How do you suppose you would find out what t1 is?
2)
Since I see you know what work is, how much work must be done on the box to make it come to rest?
3)
Do you know any calculus?
4)
You may be missing a factor of 1000 because the question uses kW = 1000 W
5)
What is the definition of force?

 

[cepheid]

For #4, I said Work/Time=Power so Fd/t=P so Ma*d/t=P because W=fd=ma*d so then I plugged in all the numbers and got .25 which leads me to believe its 250. ? I don't know I picked D

Your understanding of the physics seems fine. power = work/time = force*distance/time. Hence: P = mg*d/t

Solve for m.

I suspect that you simply made a mistake with the units. Be careful: that's 10 kilowatts, not 10 watts.

 

[girlphysics]

Let me just say, I have a terrible teacher and probably shouldn't be in AP physics, its halfway through the year and I know nothing. THANK YOU for your help. I also have not taken calculus and I have to turn this in tomorrow.
1.) t=√1/2ax ? I am not sure. I feel really stupid

 

[cepheid]

v0t+1/2at^2 so because initial is 0, just 1/2at^2??

Exactly. Now what is "a" in this situation? Hint: what is the only force that is present here?

So, you can solve for t1 using this equation. Call the initial height above the ground "h".

Then:

h/2 = 1/2*(a)*t₁²

Solving for t2 is trickier. You consider the halfway point to be the start of a new problem where you apply the same equation again. Only now there IS an initial velocity: it's equat to the velocity the object has reached at the half-way point. To compute this, you need another kinematics equation that gives you velocity vs. time.

 

[girlphysics]

OK So I got 4! yay. 1 down 4 to go.

 

[girlphysics]

a is 10m/s/s gravity...
so h= 5t^2 so t=√h/5

 

[girlphysics]

I don't know what equation..?

 

[girlphysics]

1)
How do you suppose you would find out what t1 is?
2)
Since I see you know what work is, how much work must be done on the box to make it come to rest?
3)
Do you know any calculus?
4)
You may be missing a factor of 1000 because the question uses kW = 1000 W
5)
What is the definition of force?

2.) I don't know it must only be effected by the normal force and coefficient of friction, and F=μFn. I don't know what to do.
5.) F=ma. I worked on this for like an hour. Dont I need the change in velocity? to find the impulse?

 

[girlphysics]

I just need the letter answers if its easier to work backwards...

 

[cepheid]

a is 10m/s/s gravity...
so h= 5t^2 so t=√h/5

Yeah exactly. The answer I was going for was that a = g, where g is defined as 9.81 m/s/s (or 10 m/s/s if you're not too concerned about accuracy).

Your work is pretty close. Be a little more careful with your algebra. The equation was

h/2 = (1/2)gt₁²

multiply both sides by 2:

h = gt₁²

t₁ = √(h/g)

I don't know what equation..?

You don't know how velocity varies with time during constant acceleration? Hint: what is the definition of acceleration?

The question you're trying to answer is: how fast is it going when it reaches the halfway point?

 

[SammyS]

I need help with all of these. I am so lost on all of them and have been working for hours. Anything is so appreciated.


_____1. A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. It falls the second half of the distance in t2 seconds. What is the value of t1/t2?
A) 1/2√2 B) 1/√2 C) ½
D) 1-(1/√2) E) √2-1

The wording of Question #1. must be in error. The time, t₁ to go the first half of the distance must be greater than time, t₂, the time to go the second half of the distance.

Therefore t₁/t₂ > 1 . But ALL of the answers are less than 1.

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

 

[girlphysics]

oh that makes sense!

well acceleration is the rate at which velocity changes...

v^2=v0+at

 

[girlphysics]

but there's also v^2=v0^2+a(x-x0)

 

[girlphysics]

The wording of Question #1. must be in error. The time, t₁ to go the first half of the distance must be greater than time, t₂, the time to go the second half of the distance.

Therefore t₁/t₂ > 1 . But ALL of the answers are less than 1.

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

THATS exactly what I thought at first, but I know this question is right...

 

[cepheid]

oh that makes sense!

well acceleration is the rate at which velocity changes...

v^2=v0+at

It's actually v = v0 + at. (Remember that equations have to be dimensionally consistent: if the right-hand side has dimensions of velocity, the left-hand side must as well, otherwise it is nonsense).

Can you see how to use this equation (velocity vs. time) to figure out how fast the object was going at time t = t1 when it reaches the half-way point?

THATS exactly what I thought at first, but I know this question is right...

How do you know? It could easily have a typo...

 

[girlphysics]

It's actually v = v0 + at. (Remember that equations have to be dimensionally consistent: if the right-hand side has dimensions of velocity, the left-hand side must as well, otherwise it is nonsense).

Can you see how to use this equation (velocity vs. time) to figure out how fast the object was going at time t = t1 when it reaches the half-way point?



How do you know? It could easily have a typo...

Well plug in t1 for t?
I don't know for sure, but I asked two friends who have the same problem if they were able to find the right answer and they both said yes, and I am pretty sure my teacher has been using the same problems for like his 30 years in teaching- not saying that everyone could have missed the typo.

 

[girlphysics]

how fast is it going when it reaches the halfway point? -so If I plug in t1 and use gravity, but it does have an intial velocity now right? how do I get that?

 

[girlphysics]

thank you so much cepheid for helping me- I am a dumb blonde when it comes to physics

 

[cepheid]

how fast is it going when it reaches the halfway point? -so If I plug in t1 and use gravity, but it does have an intial velocity now right? how do I get that?

You seem to be confusing the two stages of the problem. In the first stage of the problem, we consider it accelerating from rest at t = 0 to t = t1. We know that it moves a distance h/2 in this time interval, therefore we are able to solve for how long the time interval is by using the equation for distance vs. time. That's what we've done so far.

Now we want to consider the second stage, which is the stage of the motion between t = t1 and t = t1 + t2 [I EDITED THIS]. We know it moves down by another h/2 distance in this stage. So in principle, we can use the distance vs. time equation AGAIN to solve for the time interval t2 [I EDITED THIS]. The problem is, that during THIS stage, we don't know what the INITIAL velocity is (i.e. the velocity at t1).

To solve for the velocity at t1, we can use the equation for velocity vs. time: v = v0 + at, and consider it accelerating from rest at t = 0 to some final speed at t = t1. Using this equation, can you solve for how fast it is going at t = t1? Hint: what is v in this case? What is v0 in this case? What is the time interval t in this case?

Once you know the velocity at t1, you use that in the distance vs. time equation for the second stage.

 

[cepheid]

On second thought, this will be algebraically tedious. A much simpler way is to compute the TOTAL time of the motion t_total, which is the time required to move the total distance h. Then you can compute t2 using t2 = t_Total - t1.

Sorry.

 

[cepheid]

Perhaps if the time t₂ represented the time to go the total distance, then one of the listed answers might be correct.

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.

Well plug in t1 for t?
I don't know for sure, but I asked two friends who have the same problem if they were able to find the right answer and they both said yes, and I am pretty sure my teacher has been using the same problems for like his 30 years in teaching- not saying that everyone could have missed the typo.

See above.

 

[SammyS]

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.

Right.

That makes good sense.

 

[girlphysics]

so how do I solve for the total time?

 

[girlphysics]

If you assume that they want the reciprocal of what she wrote: i.e. t2/t1, then things work out in the sense that you get one of the given answers.



See above.

So do you mean that the given answer will be total time/t1? I am so confused

 

[girlphysics]

or do you mean it will turn out to just be t2/t1? Because I didnotice immediatley all the answers are less than 1 and they shouldn't be.

 

[cepheid]

so how do I solve for the total time?

EXACTLY the same way you solved for t1, with the difference that the distance traveled is the full height h, now, instead of h/2. Think about it: it's an exactly equivalent situation in both cases. You have a ball that starts at rest from some height, and then it gets dropped, and you want to know how long it will take to fall a certain distance.

or do you mean it will turn out to just be t2/t1? Because I didnotice immediatley all the answers are less than 1 and they shouldn't be.

We think that the problem is asking for t2/t1 rather than t1/t2 (which is what you wrote), simply because the former makes more sense given the answer choices (which are all less than 1).