vcsharp2003
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Since the ##PE## of stone increases when it's moved up, so net work done should be positive.rudransh verma said:How? The net force is downwards and displacement is upwards.
Since the ##PE## of stone increases when it's moved up, so net work done should be positive.rudransh verma said:How? The net force is downwards and displacement is upwards.
Well why not ##\Delta U=-W##vcsharp2003 said:Since the ##PE## of stone increases when it's moved up, so net work done should be positive.
Change in ##PE## is simply ##mgh##. What is change in ##KE##? What you've in above post is wrong since the equation you gave is the relationship between ##PE## and work done by the corresponding conservative force i.e. by mg in this case. You need to use the equation I mentioned and you need to understand the meaning of net work done.rudransh verma said:Well why not ##\Delta U=-W##
Since W=-250J so ##\Delta U= 250J## which is +ve as it should be. Meaning U increases.
Steve4Physics said:You are being asked to find the work done by L→, not the work done by W→+B→.
L=mg-rhoVg (taking g=10)Doc Al said:What's lifting the stone? That's the force that's doing the work.
I don't know about this formula. All I know is ##\Delta KE +\Delta U=0## for a closed system.vcsharp2003 said:I would use the following equation to solve this problem.
##W_{net}= \Delta KE + \Delta PE##, where ##W_{net}## is work done by applied forces (excluding force of gravity)
This equation is valid only if no forces other than gravity force act on the object. In your question, the object has an upthrust force as well as an applied force acting as the object is brought up to the surface. Therefore, your equation cannot be applied to the scenario in question.rudransh verma said:All I know is ΔKE+ΔU=0 for a closed system.
Okvcsharp2003 said:The equation that I gave is a general form of work energy theorem.
It makes problem solving easier when dealing with problems under Work,Energy and Power chapter.rudransh verma said:Ok