Physics--Tension problem?

Question

An elevator starts from rest with a constant upward acceleration and moves 1m in the first 1.5s. A passenger is holding a 9kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8m/s^2.
What is the tension in the cord as the elevator accelerates? Answer in units of N.

Answer 1

s=ut+1/2at^2
Where s is the distance travelled in meters, u is the initial velocity in m/s, a the acceleration in m/s^2, and t is the time in seconds.

Substitute given values and solve for a:

1=0*1.5+1/2a*1.5^2
1=1.125a
a=1/1.125
=0.89m/s^2

Tension in cord:

Let T = tension in cord in N
F=net force in N acting on bundle
M=mass of bundle
a=acceleration of the bundle.

The downward force exerted by the weight of the bundle is equal to 9*9.8=88.2N. Mass of the bundle is given as 9kg, and
we have calculated a to be 0.89m/s^2.

First calculate the net force given by the formula
F=Ma:
F=9*.89
=8.01N

Tension in cord=Weight of bundle +F
=88.2+8.01
=96.21N

Answer 2

The upward acceleration is given by s = .5*a*t^2. You have s and t, find a. The tension will be M*(g+a), where M is the mass of the bundle.

Answer 3

if the elevator moves upward with a constant acceleration of 0.88m/s^2 with a corresponding force of (0.88m/s^2)(9kg)= 8N and the weight of the elevator (if the passenger and the weight of the elevator itself is not included) is (9.8m/s^2)*(9kg)=88.2N so the tension is 88.2-8N= 80.2N (directed upwards)...

hehe..sorry if im wrong

Answer 4

i don't love numbers, so i visit replace the equipment weight with W, and the given perspective with theta. pressure in higher cable is T1 pressure in decrease cable is T2 The equipment isn't accelerating, so all forces might want to upload as a lot as 0. Vertical pressure stability: T1*sin(theta) = W Horizontal pressure stability: T1*cos(theta) = T2 subsequently: T1 = W/sin(theta) T2 = T1*cos(theta)/sin(theta) = W/tan(theta) files: W:=500 lbs; theta:=30 deg; outcomes: T1 = 1000 lbs T2 = 866 lbs