# Calculate the initial velocity of the elevator

1. Sep 25, 2007

### 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

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

2. Sep 25, 2007

### 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

3. Sep 25, 2007

### 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?)

4. Sep 25, 2007

### learningphysics

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

5. Sep 25, 2007

### anglum

we havent 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????

6. Sep 25, 2007

### anglum

do i give the 2.720a negative sign

and then use the formula

7. Sep 25, 2007

### learningphysics

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

8. Sep 25, 2007

### anglum

ok so i get Vi = 4.66535 m/s ???

9. Sep 25, 2007

### learningphysics

looks good.

10. Sep 25, 2007

### anglum

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

11. Sep 25, 2007

### 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??

what is kN?

do i do this one like the last problem?

12. Sep 25, 2007

### learningphysics

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.

13. Sep 25, 2007

### anglum

so i can solve it like this

X = Normal Force

X = MA + MG

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

X = 9.36kN

14. Sep 25, 2007

### anglum

or did i screw up the signs again

15. Sep 25, 2007

### learningphysics

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

also a = -1.9m/s^2

16. Sep 25, 2007

### anglum

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

17. Sep 25, 2007

### anglum

so the new equation looks like this

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

18. Sep 25, 2007

### anglum

or was the -9.8 supposed to be positive

where X would equal .6444

19. Sep 25, 2007

### learningphysics

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.

20. Sep 25, 2007

### 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