Question #303666

A 3000 kg elevator rises from rest in the basement to the floor. A distance of 25m . As it passes the fourth floor it's speed is 3 m/s . There's a constant frictional force of 500N. Calculate the work by the lifting mechanism?

Expert's answer

Explanation and Calculation.


  • The minimum needed energy or the minimum work need to be done to lift upto 25 m is,

E=mgh+fh=25 (3000×9.8+500)=29900 J\qquad\qquad \begin{aligned} \small E&=\small mgh+fh\\ &=\small 25\,(3000\times9.8+500)\\ &=\small 29900\,J \end{aligned}


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