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I have this question in my book

A bucket weighing 1.2 kg when empty is loaded with 5 kg of sand and then lifted to a height of 10 m at a constant speed. Sand leaks out of a hole in a bucket at a uniform rate. One third of the sand is lost by the end of lifting. Find the work done.

In problems like this, one assumes the use of acceleration to counter gravitational acceleration $9.8 m/s^2$.For example, to compute the work done on the bucket you have

$W=Fd = mad = (1.2kg)(9.8m/s^2)(10m) = 117.6J$

However that should give you a net force of zero and the bucket should not move upward.

A. Why can't we just assume some acceleration value $ a > g$ ? If that's the case (i.e using some acceleration value greater than g) does that mean that the actual work done on the bucket neglecting other non conservative forces, is some $W > 117.6J$ because the actual acceleration is again greater than $g$?

B. How is it possible to have a constant speed (as stated in the problem) when lifting when you have a positive acceleration?

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  • $\begingroup$ Thinking of "use of acceleration to counter gravitational acceleration" will not get you far. The solution of certain problems may suggest that interpretation, but it doesn't work as a physical principle. Think instead about forces. $\endgroup$ – garyp Jan 23 '18 at 16:16
  • $\begingroup$ The work done on the bucket is $m_b g h$. That is mass of the bucket times the force of gravity acting on the bucket times the distance it was raised. It has nothing to do with the sand. $\endgroup$ – Steven Thomas Hatton Jan 23 '18 at 16:19
  • $\begingroup$ Okay then, thinking about forces, the force done upward to counter the force done by gravity is still equal because in problems like this, one uses the same magnitude of acceleration. Hence the net force is zero, meaning no acceleration at all. Why do we always use that in solving forces or work especially involving gravity? $\endgroup$ – Panchix Regen Jan 23 '18 at 16:21
  • $\begingroup$ @StevenHatton yes, I did specify that work for the bucket in the second paragraph $\endgroup$ – Panchix Regen Jan 23 '18 at 16:23
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    $\begingroup$ Work done is not limited to the bucket. "Find the work done." You can just use the average weight. $\endgroup$ – paparazzo Jan 23 '18 at 17:56
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However that should give you a net force of zero and the bucket should not move upward.

Net force of zero implies zero acceleration, not zero velocity. If it helps, you could imagine that rather than the system starting from rest at the ground, it starts (magically) with some constant velocity and zero net force.

Practically, this is okay because we can make the acceleration arbitrarily short and the constant speed arbitrarily close to zero. That means the amount of work necessary to accelerate is also arbitrarily close to zero and can be ignored.

Why can't we just assume some acceleration value $a>g$ ?

Because most physical systems are incapable of providing either constant acceleration or constant force.

If you push a box across the room, you have to supply a bit of force to accelerate it. But once it starts moving, you can only continue to supply force while it is in contact with your hand. So the force supplied by you is variable, but sufficient to maintain a (roughly) constant speed.

As the problem has specified a constant speed, you cannot assume constant acceleration.

How is it possible to have a constant speed (as stated in the problem) when lifting when you have a positive acceleration?

That's the way motors work. If you supply a load against the motor, it can generate more torque (force). This force increases until it is larger than gravity and will accelerate the load upward. As the load accelerates, it increases speed. As the motor speed increases, torque (force) decreases until it matches the force of gravity. Acceleration falls to zero and the load rises at a (nearly) constant speed the rest of the way.

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In problems like this, one assumes the use of acceleration to counter gravitational acceleration of $9.8m/s^2$. ... However that should give you a net force of zero and the bucket should not move upward.

It is not an acceleration which is needed but an applied force to counter the weight of the bucket (ie the force on it due to gravity).

Yes, this does mean that the net force on the bucket is zero so it does not accelerate. However, this does not mean that the bucket cannot move. It can move with constant velocity, as the question states.

The bucket probably starts and ends at rest, so there must actually be initial and final brief periods of acceleration. But this does not make any difference to the amount of work done, even if the periods of acceleration were longer. For conservative forces like gravity, work done depends only on the changes in potential and kinetic energies. The answer will be the same whether the bucket begins and ends at rest, or whether it begins and ends with the same velocity.

If we apply a constant force on the bucket which is greater than its weight, the bucket accelerates and the work we do on it is greater than $mgh$. The excess work done equals the kinetic energy which the bucket acquires.

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Treat $m$ as a function of height $z$:

$$ m(z) = m_0 + \alpha z $$

with:

$$ m_0 = (1.2 + 5)\,kg = 6.2 \,kg, $$

and:

$$ \alpha = -(\frac{5}{3}\,kg)/(10\,m)$$.

Work is force times distance:

$$ W = \int_0^h{m(z)gdz}=g\int_0^h{(m_0+\alpha z)dz}$$

$$ W=(m_0+\frac{\alpha}{2}h)gh=\bar{m}gh,$$

where $\bar{m}$ is the average weight of the bucket, which is what @Paparazzi said to do.

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