While stationary on Earth you have a weight of 550 N. When in an elevator that accelerates upward, your apparent weight temporarily becomes 590 N. When descending, your apparent weight temporarily becomes 510 N. How large is the upward acceleration of the elevator

The upward acceleration of the elevator is 10.51 m/s²

Explanation:

Given;

weight on Earth while stationary, W = 550 N

apparent weight when the elevator accelerates upwards, W'₁ = 590 N.

apparent weight when the elevator descends, W''₂ = 510 N

When moving upwards;

T = ma + mg

where T is the total reading on the scale of the elevator

T = ma + W  

W'₁ = ma + W

590 = ma + 550

ma = 590 - 550

ma = 40equation (i)

where;

m is your mass

a is your acceleration

When descending;

T = mg - ma

T = W - ma

W''₂ = W - ma

510 = 550 - ma  

ma = 550 - 510

ma = 40 equation (ii)

Now, ma is the same in both equations, we calculate m first,

From Newtons second law of motion;

F = W = mg

when you are stationary on Earth, g = 9.8 m/s²

550 = 9.8 x m

m = 550/9.8

m = 56.123 kg

Finally, substitute m in any of the equations;

56.123 x a = 40

a = 40/56.123

a = 0.71 m/s²

When moving upwards;

the acceleration of the elevator becomes = (a + g)

                                                                      = (0.71 + 9.8)

                                                                      = 10.51 m/s²

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west