Calculate the initial velocity of the elevator

[anglum]

A descending elevator of mass 290kg is uniformly decelerated to rest over a distance of 4m by a cable in which the tension is 3631N

the acceleration due to gravity is 9.8m/s^2
calculate the initial velocity of the elevator at the beginning of the 4m descent

answer in units of m/s

i am confused on this one due to the introduction of velocity

 

[anglum]

using what we did in the last problem where T = MA + MG

3691 = 290A + 290(9.8)
A = 2.720

so now we have the acceleration

 

[learningphysics]

First find the acceleration using forces... then you can find the initial velocity using kinematics.

There's also a method to do it using work/energy (have you covered this yet?)

 

[learningphysics]

using what we did in the last problem where T = MA + MG

3691 = 290A + 290(9.8)
A = 2.720

so now we have the acceleration

good. now use kinematics to get the initial velocity. careful about signs.

 

[anglum]

we haven't covered work/energy

so i found the acceleration to be 2.720

so i have Vf= 0, Vi =?, A = 2.720, D = 4m

so now i can solve for Vi correct?

 

[anglum]

do i give the 2.720a negative sign

and then use the formula

Vf^2 = Vi^2 + 2AD

 

[learningphysics]

do i give the 2.720a negative sign

and then use the formula

Vf^2 = Vi^2 + 2AD

yes, use this formula. acceleration is positivie. d is negative.

 

[anglum]

ok so i get Vi = 4.66535 m/s ?

 

[learningphysics]

ok so i get Vi = 4.66535 m/s ?

looks good.

 

[anglum]

ok 2 more left if u got the time ur my saviour

 

[anglum]

A person weighing .8kN rides in an elevator that has a downward acceleration of 1.9m/s^2
the acceleration of gravity is 9.8m/s^2
what is the magnitude of the force of the elevator floor on the person??
answer in units of kN

what is kN?

do i do this one like the last problem?

 

[learningphysics]

A person weighing .8kN rides in an elevator that has a downward acceleration of 1.9m/s^2
the acceleration of gravity is 9.8m/s^2
what is the magnitude of the force of the elevator floor on the person??
answer in units of kN

what is kN?

do i do this one like the last problem?

it's very similar to the last problem... just that instead of tension, you have the normal force that the elevator exerts upward on the person. again be careful about signs.

1000N = 1kN.

 

[anglum]

so i can solve it like this

X = Normal Force

X = MA + MG

X= (.8)(1.9) + (.8)(9.8)

X = 9.36kN

 

[anglum]

or did i screw up the signs again

 

[learningphysics]

so i can solve it like this

X = Normal Force

X = MA + MG

X= (.8)(1.9) + (.8)(9.8)

X = 9.36kN

0.8kN is the weight, not the mass...

also a = -1.9m/s^2

 

[anglum]

ahh so i have to convert that weight to mass... and then solve using the equation i had used

 

[anglum]

so the new equation looks like this

X = (.081kN)(-1.9m/s^2) + (.081kN)(-9.8)

 

[anglum]

or was the -9.8 supposed to be positive

where X would equal .6444

 

[learningphysics]

so the new equation looks like this

X = (.081kN)(-1.9m/s^2) + (.081kN)(-9.8)

no...

I recommend working the whole problem in N and then converting to kN at the end...

The weight is 0.8kN = 800N. what is the mass in kg?

Also, you already know the weight (ie mg) is 0.8kN = 800N

X = mass*(-1.9) + 800N

if you are confused about signs and directions, start at the freebody diagram.

 

[anglum]

ahhh ok my conversions were off

so mass is 81.63 N

so then X = -155.1020408N + 800N

X = 644.89795 N

X = .64489795 kN

 

[anglum]

ok i got that one

one last problem

Take the mass of the Earth to be 5.98*10^24 kg.
If the earhts gravitation force casues a falling 65kg student to accelerate downward at 9.8m/s^2, determine the upward acceleration of the Earth during the students fall. answer in units of m/s^2

 

[learningphysics]

ahhh ok my conversions were off

so mass is 81.63 N

so then X = -155.1020408N + 800N

X = 644.89795 N

X = .64489795 kN

yeah, that looks right.

 

[learningphysics]

ok i got that one

one last problem

Take the mass of the Earth to be 5.98*10^24 kg.
If the earhts gravitation force casues a falling 65kg student to accelerate downward at 9.8m/s^2, determine the upward acceleration of the Earth during the students fall. answer in units of m/s^2

What is the force the Earth exerts on the student? What is the force the student exerts on the earth?

 

[anglum]

this last problem i just posted is it similar in how i solve as the last one?

 

[anglum]

force from Earth is = MA + MG
but i don't have A

force for student is MA + MG
65*9.8 + 65*9.8

 

[learningphysics]

this last problem i just posted is it similar in how i solve as the last one?

no, this one is more conceptual.

 

[anglum]

are my formulas for the force of each right or no

 

[learningphysics]

force from Earth is = MA + MG
but i don't have A

force for student is MA + MG
65*9.8 + 65*9.8

no... it's not like that... don't get caught up in just using ma + mg blindly... that only worked in very specific cases. always draw a freebody diagram first... better not to take shortcuts... it will lead to mistakes.

 

[anglum]

so the acceleration for the student is 9.8m/s right? and his mass is 65kg
and i have the Earth's mass but not the acceleration

 

[learningphysics]

so the acceleration for the student is 9.8m/s right? and his mass is 65kg
and i have the Earth's mass but not the acceleration

Yes... what is the force the Earth exerts on the student?

 

[anglum]

i have the ma+mg stuck in my head. and drawing it isn't helpin me in explaining to u where i am

 

[learningphysics]

It is always best to start your force problems by writing out the equations in the form:

$$\Sigma{F} = ma$$

and going from there... don't take shortcuts.

 

[anglum]

the force the earht exerts is

x = MA
x = 65*9.8
x=637N

 

[learningphysics]

i have the ma+mg stuck in my head. and drawing it isn't helpin me in explaining to u where i am

don't get stuck with the ma + mg... always start from the basics...

$$\Sigma{F} = ma$$.

But anyway, for this problem, what is the force the Earth exerts on the student?

 

[learningphysics]

the force the earht exerts is

x = MA
x = 6589.8
x=637N

yes. So what is the force the student exerts on the earth?