Conservation of engery and work

[Ereny]

a partially filled bag of cement is tossed off of bridge with an initial upward velocity of 3.0 m/s. the bag has a mass of 16 kg and falls 40.0 m to the river below. determine the velocity of the bag just before striking the water.
i found that and it comes out to be 28.2 m/s ( correct me if i am wrong please)

b. imagine that water were replaced with a trampoline . the trampoline has a spring constant of 2000 N/m. determine the distance the bag will sink in the trampoline before coming to a stop.?

i know that i have to use kinetic energy and potential energy and elastic potential energy but i don't know what to set = to zero and what not..
can someone please help me !

 

[tiny-tim]

Hi Ereny!

i know that i have to use kinetic energy and potential energy and elastic potential energy but i don't know wt to set = to zero and what not..

you don't have to set anything to zero

for potential energy, all that matters is the difference between PE before and PE after

 

[Ereny]

and the equation for PE is 1/2kx^2 right?

 

[tal444]

That is the equation for the PE of a spring, correct. Not necessarily PE in general.

 

[Ereny]

i did that but i got a really big number for the distance i got the same distance as its given..

i did : 1/2(2000)x^2 - 1/2(2000)(40)^2=0 ? i don't think this is right

 

[berkeman]

a partially filled bag of cement is tossed off of bridge with an initial upward velocity of 3.0 m/s. the bag has a mass of 16 kg and falls 40.0 m to the river below. determine the velocity of the bag just before striking the water.
i found that and it comes out to be 28.2 m/s ( correct me if i am wrong please)

I think you left off the initial downward velocity that the bag has as it passes you on its way down...

b. imagine that water were replaced with a trampoline . the trampoline has a spring constant of 2000 N/m. determine the distance the bag will sink in the trampoline before coming to a stop.?

i know that i have to use kinetic energy and potential energy and elastic potential energy but i don't know what to set = to zero and what not..
can someone please help me !

i did that but i got a really big number for the distance i got the same distance as its given..

i did : 1/2(2000)x^2 - 1/2(2000)(40)^2=0 ? i don't think this is right

The 2nd term in your equation looks wrong. When the bag initially hits the trampoline, it has kinetic energy. When it is stopped by the springs of the trampoline, it has potential energy stored in the springs, and zero kinetic energy. There is also a small difference in gravitational PE between the position of the bag when it just hits the trampoline, and when it is stopped at a somewhat lower position...

 

[Ereny]

okay so here's my work correct me if i am wrong please

change KE + change PE + change GPE = 0
F(1/2mv^2)-I(1.2mv^2) + F(1/2kx^2)- I(1/2kx^2) + F(mgh)-I(mgh) =0
(1/2*16kg * 28.2 - 0) + (0-1/2 * 2000N/m * X^2 ) + (0- 16kg * 9.81* 40m) = 0

i solved for X and i got a neg # idk why but the number is kinda right because that's what the teacher kind off had..
did i mess something in my work that's why i got a neg # idk...

 

[Ereny]

i think i see where i messed up i forgot to square the 28.2 .. but when i did i got X= .29 m
does that look okay? or is it too small...

 

[berkeman]

okay so here's my work correct me if i am wrong please

change KE + change PE + change GPE = 0
F(1/2mv^2)-I(1.2mv^2) + F(1/2kx^2)- I(1/2kx^2) + F(mgh)-I(mgh) =0
(1/2*16kg * 28.2 - 0) + (0-1/2 * 2000N/m * X^2 ) + (0- 16kg * 9.81* 40m) = 0

i solved for X and i got a neg # idk why but the number is kinda right because that's what the teacher kind off had..
did i mess something in my work that's why i got a neg # idk...

i think i see where i messed up i forgot to square the 28.2 .. but when i did i got X= .29 m
does that look okay? or is it too small...

I see several problems. The initial downward velocity of the object as it passes you on the way down is not zero. What is it? The final velocity of the object is not 28.2m/s when it is stretching the trampoline. It is zero there. You are mixing the two conditions together (right before the object hits the trampoline, and when it is stopped farther down by the trampoline). You are also mixing up the 40m with the delta-h from the bridge to the position that the object stops in when the trampoline is extended downward.

You should be more careful to define the conditions at each step in the sequence:

-1- When the object is just about to get thrown upwards (PE, KE)

-2- When the object reaches the top of its arc (although this is not really needed)

-3- When the object passes by you on the way down

-4- Right before when the object hits the resting trampoline

-5- When the object is stopped by the trampoline at the bottom.

.

 

[Ereny]

would the initial velocity be 3 then.. ahh i am soo confused and oohh my teacher said to ignore the part where the trampoline goes downward and that way it will be easier to figure out i guess.

 

[tiny-tim]

Hi Ereny!

(just got up :zzz:)

would the initial velocity be 3 then.. ahh i am soo confused and oohh my teacher said to ignore the part where the trampoline goes downward and that way it will be easier to figure out i guess.

the initial kinetic energy is your original 1/2 m 28.2² (actually, i got a very slightly different figure)

the final kinetic energy is 0

the initial elastic potential energy is 0

the difference in gravitational potential energy (in the height of the surface of the trampoline) is small, and your teacher is telling you to ignore it (but don't do that in an exam unless the question tells you to )