Question:
An elevator of mass 500 kg descends with an acceleration of 2 m/s. If the lift descends 10 m under these circumstances, find the work done by the tension m in the lift cable
Answer:
Let the tension in the cable be T
The equation of motion for the lift is:
T−mg=ma where m=mass of the lift, a= acceleration of the lift)
⇒T=m(g+a)
⇒T=500(10+2)=500×12=6000N.
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