Question

A lift is going up. The total mass of the lift and passenger is 1500kg. the variation of the speed of the lift is given below as function of timea. Identity the region where the lift is in uniform motion b. What is the height to which the lift takes the passengers? c. What will be the tension in the rope at t=1sec and t=11sec? ​

Answer 1

The height attained is 36 m and the tension in the rope is 12000 N.

Explanation:

Given data:

m = 1500 Kg

To find

Height "h"= ?

Tension "T" = ?

(a)  Area under u - t curve gives distance covered ( Height)

     Height attained = 1/2 (12 + 8) x 3.6

                                = 20 x 1.8

                                = 36 m

(b) Tension

T = mg + ma => T = m (g +a)

T = 1500 ( 9.8 + 1.8)

T = 17400 N

• At t = 6 seconds T is in upward direction and mg in downward direction

V = constant , a = 0

T = mg = 1500 x 9.8 = 14700 N

• At t = 11 second

a = 3.6 m/s , v = 0

a = v - u / t = 0 - 3.6 / 2 = - 1.8 m/s^2 ( Retardation)

T = m (g +a)

   = 1500 (9.8 - 1.8) = 1500 x 8 = 12000 N

Answer 2

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