Power mechanics physics problem

AI Thread Summary
To match a power output of 200 W while climbing a 5.0 m rope, the student must calculate the work done using the formula Work = mgh, where m is mass and h is height. The power formula indicates that Power = Work/time, allowing the determination of time required to achieve the desired power output. By rearranging the equations, the average speed can be derived as Average speed = height/time. The discussion emphasizes the need to find the correct time to complete the climb to meet the power requirement.
-Aladdin-
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A 50kg student climbs a 5.0m long rope and stops at the top.What must her average speed be in order to match the power output of 200w.
a)0.20m/s
b)0.41m/s
c)0.10m/s

My Work :
And stops at the top => v=0 and h=5m so I'll use mgh.
Now power output = variation of energy/1

But average means the variation of x /variation in time.

Any help will be great,
Thanks
 
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How much work does she have to do to climb the rope? In what time must she complete the climb to have her average power equal 200 W?
 


Doc Al said:
How much work does she have to do to climb the rope? In what time must she complete the climb to have her average power equal 200 W?

Work = mgh , where h=5

Power = Work/time

time = Work / power

Average speed = 5 / time(gotten) ?
 


Sounds good to me.
 
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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