Question

a student stands on a scale in an elevator that is moving upward and accelerating at 2.2 m/s^2. if the student has a mass of 57 kg to the nearest N what is the scale reading?

Answer 1

with these type of problems I would draw a picture first then a free body diagram of the stufdent in an elevator

From the free body diagram you use \[\Sigma F = ma\] and solve for the force pushing up on the student

Answer 2

ok I did that and I add the forces and I got 684.5 N... Do that right?

Answer 3

Yes that should be the answer assuming A = +2.2 m/s^2 upward

Answer 4

Can you help me to this problem:
A skier traveling at 29.6 m/s encounters a 20.5 slope. If you could ignore friction, to the nearest meter, how fair up the hill does go?

Answer 5

Do you mean a 20.5 degree slope? like this picture:

Answer 6

yes

Answer 7

and the initial velocity is 29.6... and they what me to find the distance... however I don't how to find the acceleration

Answer 8

Again I would draw a picture and place the motion variables (x,v,t, and a) into it


Do you have the mass of the skier? or the time?

Answer 9

no

Answer 10

Ok you can solve this system does the above picture make sense? we can set the final position Xf = D the distance traveled

You will have to use these three equations:

\[V_f^2 = V_i^2 + 2(-a)D\]
\[D = V_it_f + \frac{1}{2} (-a)t_f^2\]
\[V_f = V_i + (-a)t_f\]

you need to solve this system for D. The three unknowns are D,a,tf

Answer 11

The middle equation has D by iteself (which you are solving for)

Find tf in terms of -a with the third equation

Find -a in terms of D with the top equation

Answer 12

yes and there is where I got stuck...because I don't how to find the other unknown... :(

Answer 13

You dont need to find the time or acceleration to solve...ok can you find tf by itself from the third equation first?

Answer 14

and the acceleration, because if I put in the second equation I will have unknown d and -a...

Answer 15

Yes then you use the first equation to find (-a) in terms of d and solve for d then

Answer 16

ok... I did and I got 22.2m do that right?

Answer 17

when I did solve for d I got this equation:
1/2d=vi +1/2(vf-vi)

Answer 18

I am solving this system and I have D dropping out I dont think this path will work because the top equation is not unique it is a combination of the others.... sorry I dont see a path if you dont have the mass of the skier

Answer 19

ok.. thanks... nut what about this problem:
a 23 kg child sits on a 5 kg sled and slides down a 104m, 29 slope, to nearest m/s what is his speed at the bottom?

Answer 20

Ok this one is solvable draw a picture:

Answer 21

yes...

Answer 22

You are solving for Vf

(1) you need to draw a free body diagram for the child in the starting and find the component of gravity down the ramp (this force will be the cause of the acceleration)

(2) Use \[\Sigma F = ma\]
to find a

(3) use \[V_f^2 = V_i^2 + 2aD\] to find the final velocity

Answer 23

This is assuming a frictionless ramp?

Answer 24

I guess ... it doesn't say....do you add the masses? (23kg+5kg)?

Answer 25

Yes

Answer 26

how do you find the other forces because in my body diagram I have for forces gravitational force and Normal force \[\Sigma F\]

Answer 27

Like this diagram:

Answer 28

yes

Answer 29

this ok?