Yes, that is correct. Good job!

[Unkownentity]

From a height of 6,00m from the floor a man drops a package weighing 1,3kg. Another man below at the height of 1,60m from the floor stops the package 0,30m from the floor. (We take for granted that the man uses a constant amount of Newton on the package and that there is no air resistance)

A) Determine the velocity of the package when the man stops the package
B) Determine the acceleration of the package during the deceleration
C) Determine the amount of Newtons the man uses on the package during the deceleration

Is there anyone who can help me out with this question showing every step, been sitting with it a while now and don't know anywhere else to turn =/

Thank you in beforehand :)

 

[PhanthomJay]

Hi, welcome to PF! Someone here would be glad to help, but Forum Rules first require that you list what you think are the relevant equations and then show some attempt at a solution. Note that the low guy stops the package over a distance of 1.6 -0.3 = 1.3 m, by moving his hands down over that distance as he catches the package. You can use energy methods if you are familiar with them.

 

[rock.freak667]

Think of conservation of mechanical energy for the first part.

You will also need the kinematic equations of motion for the second part.

 

[tiny-tim]

Welcome to PF!

Hi Unkownentity! Welcome to PF!

Try A) first: you have the initial velocity, and the total distance, and you want the final velocity, so what formula or principle do you think you should use?

EDIT: ooh, PhanthomJay and rockfreak667 both beat me to it!

 

[Unkownentity]

This is as far as i have come

By using Newtons second law F = M * A

and the acceleratoion in this case is the gravity therefore i cna rewrite the formula F = M * G

F = 1,3 * 9,82
F = 12,76

(i'm bad at using symbols over computer, what i mean with V0 is the starting velocity which should be zero in this case)

what i was thinking is by using this formula s = V0 *t * 1/2at^2 shouldn't i be able to get the amount of time it takes for it to fall and then by using (After getting the time) v = V0 + at i can find out the velocity, does that sound reasonable?

 

[tiny-tim]

Hi Unkownentity!

(you don't need the force … the acceleration is enough )

what i was thinking is by using this formula s = V_{0 * t * 1/2at² shouldn't i be able to get the amount of time it takes for it to fall and then by using (After getting the time) v = v0 + at i can find out the velocity, does that sound reasonable?}

<sub>Yes, that's the right idea.

But you only seem to know two of the standard constant acceleration equations …

there is a third one, v² = v0² + 2as</sub>

 

[Unkownentity]

Yeah isn't that the timless formula or something we haven't got to learn that yet since that goes more in depth in the next course :P but otherwise my other idea was right?

 

[tiny-tim]

Yeah isn't that the timless formula or something we haven't got to learn that yet since that goes more in depth in the next course :P but otherwise my other idea was right?

(it's not the tiny-timless formula )

Yes, those two equations together will work fine.

ok, next try B) … you have the initial and final velocity, and the distance, and you want the acceleration.

 

[Unkownentity]

i got that the answer on A was 10,50 m/s

V0 = 0
V1 = 10,5 m/s

since we only want to know the acceleration during the deceleration should i instead use the distance 0,3m? then i could use S = 1/2 (V0 + V1) *t to get the time and then use S = V0 *t + 1/2at^2 and leave the acceleration unknown does this work?

 

[tiny-tim]

You're not reading the question properly …

the package does not stop at the floor, it stops at 0.30m (and the deceleration started at 1.60m).

But your method is correct: you can use the same equations "backwards" as in A), because the same four variables are involved.

 

[Unkownentity]

Oh okey so i just need to switch the distance, so i use 1,6 instead of using 0,3?

 

[tiny-tim]

Oh okey so i just need to switch the distance, so i use 1,6 instead of using 0,3?

In part A), yes.

 

[Unkownentity]

shouldn't the distance on part A be 5,70 meters since the man stops it at that height?

 

[tiny-tim]

shouldn't the distance on part A be 5,70 meters since the man stops it at that height?

hmm … it's a badly-worded question …

From a height of 6,00m from the floor a man drops a package weighing 1,3kg. Another man below at the height of 1,60m from the floor stops the package 0,30m from the floor. (We take for granted that the man uses a constant amount of Newton on the package and that there is no air resistance)

A) Determine the velocity of the package when the man stops the package

Part A) can't mean what it says (ie at 0.30m), because the velocity when he stops it is obviously zero , so the first part must mean that he starts stopping it at 1.60m, and stops stopping it at 0.30m … and A) is asking for the velocity when he starts stopping it.

 

[Unkownentity]

i think he wrote it wrong i think he means that

A) Determine the velocity of the package when the man grabs the package

 

[tiny-tim]

i think he wrote it wrong i think he means that

A) Determine the velocity of the package when the man grabs the package

Yes, at 1.60m.

 

[Unkownentity]

Okey so i got a veloity of 0,59m/s on A) does that sound reasonable?

But what i don't understand is what distance I'm supposed to use on B) =/

 

[tiny-tim]

Okey so i got a veloity of 0,59m/s on A) does that sound reasonable?

How do you get that?

But what i don't understand is what distance I'm supposed to use on B)

1.60m - 0.30m = 1.30m.

 

[Unkownentity]

V = X
A = 9,82
t = 0,06

V = V0 + at by incorperating the numbers into the formula i got V = 0,59m/s
----------------------------------------------------------------
S = 1,6m
V0 = 0m/s
t = X

S = V0 *t + 1/2at^2

Here i did the same thing but with "T" so i could ge the time. (T = 0,06s)

thats how i did it

 

[tiny-tim]

Hi Unkownentity

(just got up :zzz: …)

V = X
A = 9,82
t = 0,06

V = V0 + at by incorperating the numbers into the formula i got V = 0,59m/s
----------------------------------------------------------------
S = 1,6m
V0 = 0m/s
t = X

S = V0 *t + 1/2at^2

Here i did the same thing but with "T" so i could ge the time. (T = 0,06s)

thats how i did it

where does 0.06 come from?

where does 1.6m come from?

 

[Unkownentity]

Question:

From a height of 6,00m from the floor a man drops a package weighing 1,3kg. Another man below at the height of 1,60m from the floor stops the package 0,30m from the floor. (We take for granted that the man uses a constant amount of Newton on the package and that there is no air resistance)

A) Determine the velocity of the package when the man grabs the package
B) Determine the acceleration of the package during the deceleration
C) Determine the amount of Newtons the man uses on the package during the deceleration

Answer:

A) S = V0*t + 1/2at^2
1,6 = 0,5(9,82) * t^2
1,6 = 4,91t^2
0,56 = t

V = V0 + at
V = 0 + 9,82 * 0,56
V = 5,5 m/s
The velocity was: 5,5m/s

B) S = 1/2(V0 + V1) *t
S = V0 * t + 1/2at^2
1,3 = 1/2(5,5) * t
1,3 = 5,50t
0,23 = t

1,3 = 1/2(a) * 0,23^2
1,3 = 0,5a * 0,05
26 = = 0,5a
13 = a

Acceleration was: 13m/s^2

C) F = ma (Newtons second law)
F = 1,3 * 13
F = 16,9N

Force was: 16,9N

Is there anywhere i have done wrong? if so where?

 

[tiny-tim]

Hi Unkownentity!

(please use the X² and X₂ tags just above the Reply box )

A) S = V0*t + 1/2at^2 V = V0 + at
1,6 = 0,5(9,82) * t^2 V = 0 + 9,82 * 0,56
1,6 = 4,91t^2 V = 5,5 m/s
0,56 = t
The velocity was: 5,5m/s

(why are you writing two separate things on the same line?? solve one equation first, then solve the next equation underneath it)

No, you're using s = 1.6m, you need to use the difference in height, which is 4.4m.

B) S = 1/2(V0 + V1) *t S = V0 * t + 1/2at^2
1,3 = 1/2(5,5) * t 1,3 = 1/2(a) * 0,23^2
1,3 = 5,50t 1,3 = 0,5a * 0,05
0,23 = t 26 = = 0,5a
13 = a
Acceleration was: 13m/s^2

Sorry, but S = 1/2(V0 + V1) *t is not one of the constant acceleration equations.

You need to use V = V₀ + at, same as in part A).

 

[Unkownentity]

hmm okey but if i use the formula you showed before V^2 = V0^2 + 2as

5.5^2 = 2a * 2,6
30,25 = 2a * 2,6
11,63 = 2a
5,81 = a

acceleration is 5,81 m/s^2

This should be correct right?