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A weight lifter lifts a mass of 250 kg with a force 5000 N to the height of 5 m. (a) What is the work done by the weight lifter! (b) What is the work done by the gravity? (c) What is the network done on the object? |
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Answer» Solution :(a) When the weight lifter lifts the mass, force and DISPLACEMENT are in the same direction which means that the angle between them `THETA= 0^(@)`. Therefore, the work done by weight lifter `w_("weight lifter")=F_(u)hcostheta=F_(w)h(coso^(@))` `=5000xx5xx(1)=25,000"joule"=25KJ` (b) When the weight lifler lifts the mass, the gravity acts downwards which means that force and displacement are in OPPOSITE direction. Therefore, the angle between them `theta=180^(@)` `w_("gravity")=F_(g)hcostheta=mgh(cos180^(@))` The net work done (or total work done ) on the object `w_("net")=w_("weight lifter")+w_("gravity")` `=25KJ-12.5KJ=+12.5KJ` |
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