Question

A block of mass m = 10 kg hangs from ceiling of
an elevator, through a string, as shown in the
figure. If the lift moves up with a constant speed
10 m/s, then the work done by tension in string
on mass m, in 2 s w.r.t. an observer on the
ground will be (g = 10 m/s2]​

Answer 1

Answer:

-2000 J

Explanation:

$as \: it \: moves \: upward \: with \: constnt \: speed \: \\ accleration \: is \: = 0 \\ \\ tension = mg = 100newton \\ \\ workdone = mgh \\ h = - \frac{1 }{2} gt {}^{2} \: \: \: \: \: \: (t = 2s) \\ h = - 20m \\ \\ negative \: sign \: shows \: downward \: direction \\ as \: positive \\ w = mgh = 100 \times - 20 \\ = - 2000 \: joule \\ \\ negative \: sign \: shows \: that \: displacement \\ is \: not \: in \: the \: dirctionof \: force$

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Answer 2

Answer:

2kj

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