Q130 / 160 · TS EAMCET 2020 (11 Sep Shift 1) · Physics

An elevator of mass $500 \mathrm{~kg}$ is ascending upwards with a constant acceleration $a=2 \mathrm{~m} / \mathrm{s}^2$. What is the work done by the tension in the elevator cable during its climb by $12 \mathrm{~m}$ ? (Take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

  • A
    $36 \mathrm{~kJ}$
  • B
    $48 \mathrm{~kJ}$
  • Correct answer
    C
    $72 \mathrm{~kJ}$
  • D
    $100 \mathrm{~kJ}$

Solution

Mass of elevator, $m=500 \mathrm{~kg}$ Acceleration of elevator, $a=2 \mathrm{~ms}^{-2}$ (upward) When elevator is going upward, then tension in the elevator cable, $T=m(g+a)=500(10+2)=6000 \mathrm{~N}$ $\therefore$ Work done by the tension force in the cable, $W=T \times s=6000 \times 12[\because \text { given } s=12 \mathrm{~m}]$ $=72000 \mathrm{~J}=72 \mathrm{~kJ}$

Hence, the correct answer is option: 'C'.

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