Work Equals Potential and Kinetic Energy - Clarification Needed

AI Thread Summary
Work is not directly equal to potential energy (PE) or kinetic energy (KE); rather, it represents the conversion of energy from one form to another. The relationship between PE and KE is contingent on the absence of external forces like friction, which can alter the energy transfer. When work is done on an object, the loss of potential energy can equal the gain in kinetic energy only in ideal conditions. In scenarios with friction, the kinetic energy gained will be less than the potential energy lost due to energy dissipation. Therefore, the statement that work equals potential energy and kinetic energy is an oversimplification and not universally applicable.
moomoocow
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hi!

can i say that
when work is done, it is equal to potential energy, which is equal to kinetic energy?

can somebody please correct me if i misunderstood it?
 
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nah u can't say that.basically u can't even say PE is equal to KE its the change that is equal
 
when work is done energy is converted from one form to the other

the potentail energy may not necssarily equal the kinetic energy... what if there was friction or some other external force?
 
oo
so its the loss of potential energy which is equal to KE?
 
moomoocow said:
oo
so its the loss of potential energy which is equal to KE?

this is not always true

think of this:

suppose there was box on top of a hill. And the surface of the hill had friction. Then as the box slides down the hill what can you say about the kinetic energy gained by this box? Is it equal to the potnetial energy lost? Wh or Why not??
 
suppose u push a box on a frictionless surface, PE wrt the floor will always be zero but there is still some KE
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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