Question

A 590 kg elevator accelerates upward at 1.2 m/s^2for the first 15 m of its motion. How much work is done during this part of its motion by the cable that lifts the elevator? Neglect any friction.

Answer 1

WD = 99120 J

Step-by-step explanation:

Elevator is connected by a cable,

(i) mg is downward,

(ii)Tension, T is upward and

(iii) acceleration, a is also upward

So, T is positive and mg is negative. We get equation,

T - mg = ma

T=m(g+a)

T = 590(10 + 1.2) =590(11.2)

T = 6608 N

As work done = Force X Displacement

Consider, 'T' as force,

WD = 6608 X 15 = 99120 J