What is the tension in the cord as the elevator accelerates?

[runner1738]

elevator tension problem

An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1.5s. A passenger in the elevator is holding a 9.8 kg is holding a 9.8 kg bundle at the end of a vertical cord. G=9.8 m/s^2. What is the tension in the cord as the elevator accelerates? Answers in units of N.

my work: (9.8kg * g)((4/9 m/s^2)/g + 1) = 96.04(1.045351474)=100.39555556 the 4/9 is the acceleration in the y direction 1/1.5/1.5 so m/s^2 right? this wasnt right what is wrong?

 

[runner1738]

t=MAy+ mg? so I am guessing the thing i have wrong is acceleration right?

 

[runner1738]

the answer was 104.751 someone said a=v^2/2d then use a in ma+mg but that never gave me 104.751 unless i just can't add, anyone?

 

[admajoremdeigloriam]

well first you need to find the acceleration of the elevator

x_f=x_i + v_i + 1/2 at²

1m=0 + 0 + .5a(1.5)² so a= 1m/(.5 x 2.25s²)

this will be vectorally added with the acceleration due to gravity
then tension will be equal to the weight T=weight=m(g+a)

T=9.8kg x (9.8 m/s² + 1m/.5 x 2.25s²)

that'll give you exactly 104.7511111 N

just remember in these cases, the elevators acceleration just adds to gravity
if the elevator was heading down you need to subtract it