Why Is Speed Proportional to the Square Root of Time in This Physics Problem?

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The discussion revolves around a physics problem where a particle's speed is questioned in relation to time. Initially, there is confusion regarding whether speed is proportional to time or the square root of time. The key clarification is that the net force acting on the particle is constant, while the rate of work done is proportional to time. This distinction leads to the conclusion that speed is indeed proportional to the square root of time, as derived from the relationship between power, work, and kinetic energy. Understanding the difference between force and the rate of work is crucial to solving the problem correctly.
NewtonGalileo
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Homework Statement



A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:

sq root of (t)

t

t^2

1/sq root(t)

1/t


Homework Equations



power = work / time
work = change in kinetic energy

The Attempt at a Solution


So,
(1/2 mv^2 / t) proportional to (t)
So,
v proportioanl to t

But, the answer key says v proportional to sq root of t. How is that?
 
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NewtonGalileo said:
power = work / time
work = change in kinetic energy

The Attempt at a Solution


So,
(1/2 mv^2 / t) proportional to (t)
Be careful here. Power is also equal to force times velocity.

In this particular problem, the force is proportional to t, as you know.

But the power is also proportional to velocity. But what is velocity?

Make the substitutions into force and velocity, and the answer should become more clear.
 
But,

net force is a constant. It is not proportional to time t.
I also looked at:
power = work/time = force x velocity
power ~ t
force x velocity ~ t
since force is a constant,
velocity ~ t

wouldn't this be right? why sq.root of t?
 
Perhaps you're right. I think I originally misinterpreted the problem statement as saying the force was proportional to t. But now I see it says that the rate of work is proportional to t. So yeah, that makes a difference.
 
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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