An elevator cab that weighs 27.8 kN moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of 2 1.22 / msand (b) decreasing at a rate of 2 1.22 / ms?

a)T = 8.63  × 10 ⁴ N, b)T = -3.239 × 10 ⁴ N

Explanation:

Given:

W = 27.8 KN = 27.8 × 10 ³ N,

For upward motion: Fnet is upward, Tension T is upward and weight W is downward so

a) a=21.22 m/s² ( not written clearly the unit. if it is acceleration?) then

Fnet = T - W

⇒ T = F + W = ma + W

T = (W/g)a + W                           (W=mg ⇒m=W/g)

T = (27.8 × 10 ³ N / 9.8 ) 21.22 m/s² + 27.8 × 10 ³ N

T = 86,263.265 N

T = 8.63  × 10 ⁴ N

b) For Declaration in upward direction a = -21.22 m/s²

Fnet = T - W

⇒ T = F + W = ma + W

T = (W/g)a + W                        

T = (27.8 × 10 ³ N / 9.8 ) (-21.22 m/s²) + 27.8 × 10 ³ N

T = -32395.5 N

T = -3.239 × 10 ⁴ N

as Tension can not be negative I hope the value of acceleration and deceleration is correct.

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.

he made 11 calls in one day