# An elevator with a wieght of 27.8 kN is given an upward acceleration of 1.22 m/s^2 by a cable. a) what is the tension in the cable. b) what is the tension when the elevator is declerating at the rate of 1.22m/s^2 but is still moving upwards?

what exactly is a kN? to find the tension of part a what equation would I use? for b) I need to make a=-1.22m/s^2 but I am not sure what the equation I would use be.

9 months ago

## To better understand the problem, let's break it down step by step:

1) What is a kN?

The notation kN stands for kilonewton, which is a unit of force commonly used in engineering and physics. One kilonewton is equal to 1000 newtons (N). So, 27.8 kN is equivalent to 27,800 N.

2) What equation can be used to find the tension in the cable?

a) To find the tension in the cable when the elevator is accelerating upwards, you can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a). The equation is:

F = m * a

In this case, the force is the tension in the cable, which we need to find. The mass (m) is not provided, but we can calculate it using the weight (W) and the acceleration due to gravity (g). The equation is:

W = m * g

To solve for m, we can rearrange the equation:

m = W / g

Substituting the known values and solving for m:

m = 27,800 N / 9.8 m/s^2 â‰ˆ 2837 kg

Now, we can substitute the calculated mass into the first equation to find the tension (T):

T = m * a
T = (2837 kg) * (1.22 m/s^2)
T = 3459.14 N â‰ˆ 3.46 kN

Therefore, the tension in the cable when the elevator is accelerating upwards is approximately 3.46 kN.

3) What equation can be used to find the tension in the cable when the elevator is decelerating?

b) When the elevator is decelerating but still moving upwards, we need to consider the net force acting on it. In this scenario, the net force should be the difference between the weight force (W) and the force required to decelerate the elevator (F_deceleration).

Since the elevator is still moving upwards, the weight force remains the same as in part a), which is 27,800 N.

To calculate the force required to decelerate the elevator, we use the equation:

F_deceleration = m * (-a)

Where m is the same mass as calculated in part a), and a is the deceleration rate of 1.22 m/s^2 but now with a negative sign to indicate the opposite direction.

Substituting the values:

F_deceleration = (2837 kg) * (-1.22 m/s^2)
F_deceleration = -3459.14 N â‰ˆ -3.46 kN

Since the tension in the cable is the net force, we can calculate it by subtracting F_deceleration from the weight force:

T = W - F_deceleration
T = (27,800 N) - (-3459.14 N)
T â‰ˆ 31,259.14 N â‰ˆ 31.26 kN

Therefore, the tension in the cable when the elevator is decelerating at a rate of 1.22 m/s^2 but still moving upwards is approximately 31.26 kN.