# Homework Help: Minimum Power for Lifting Sand

1. Oct 18, 2007

### odie5533

1. The problem statement, all variables and given/known data
A sand mover at a quarry lifts 2,000kg of sand per minute a vertical distance of 12 meters. The sand is initially at rest and is discharged at the top of a sand mover with speed 5m/s into a loading chute. At what minimum rate must power be supplied to this machine?
A) 524w B) 1.13kw C) 4.34kw D) 3.92kw E) 6.65kw

3. The attempt at a solution
$$W_{MS} = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{0}^2$$
$$W_{MS} = \frac{1}{2}(2000)(5)^2 - \frac{1}{2}(2000)(0)_{0}^2$$
$$W_{MS} = 25000J$$

There wasn't a picture, and I get kinda lost after this point.

2. Oct 18, 2007

### malty

You will definitely need to draw diagram. Is that the total amount of work that the sand mover does???

Hint *how did the sand get up there in the first place*

3. Oct 18, 2007

### odie5533

$$W_{MS}$$ is the Work of the Mover on the Sand.

A Force moved it up.
$$\sum F = F_{MS} - F_{w}$$
$$F_{w} = mg = (2000kg)(g) = 19613.3N$$
$$\sum F = F_{MS} + (-19613.3N)$$
$$W_{total} = Fd = F_{MS}d + (-19613.3N)d$$
$$W_{total} = 25000J$$ From last post
$$25000J = d(F_{MS} + (-19613.3N))$$
$$d = 12m$$ given
$$2083 = F_{MS} + (-19613.3N)$$
$$F_{MS} = 21696N$$

I'm lost =/

4. Oct 18, 2007

### odie5533

Here's my second try:
$$weight = w = mg = 2000 * g$$

Work of the Weight of the Sand
$$W_{WS} = wd = (g)(2000)(12) = 2.35 x 10^5 J/min$$

Work of the Mover on the Sand:
$$W_{MS} = \frac{1}{2}mv_{f}^2 = \frac{1}{2}(2000)(5^2) = 2.5 x 10^4 J/min$$

$$P_{k} = \frac{2.5 x 10^4}{60 s} = 416.7 W$$
$$P_{grav} = \frac{2.35 x 10^5}{60 s} = 3924 W$$
$$P_{total} = 4.34 kw$$

It seems wrong still. The mover moves 2000kg of sand 12m in 60 seconds, or 0.2m/s. How can it also move at 5m/s?

5. Oct 19, 2007

### qspeechc

Work is change in energy. So calculate the initial energy the sand has (zero), and subtract it from the total energy it has in the end, it will have both kinetc and potential energy. Then you have the work. Then power is just rate of work.

6. Oct 19, 2007

### odie5533

Another attempt:
$$W_{total} = K_{f} + PE - K_{0}$$

$$K_{0} = \frac{1}{2}mv_{0}^2$$
$$K_{0} = \frac{1}{2}(2000)(0)^2 = 0$$

$$K_{f} = \frac{1}{2}mv_{f}^2$$
$$K_{f} = \frac{1}{2}(2000)(5)^2 = 25 kJ$$

$$PE = mgh = (2000)(9.8)(12) = 235 kJ$$

$$W_{total} = K_{f} + PE - K_{0} = 25 + 235 - 0 = 260 kJ$$

$$P_{total} = \frac{W_{total}}{time} = \frac{260 kJ}{60 s} = 4.33 kw$$

Same answer as before, except this time it made a bit more sense in my mind. I still don't understand how the sand can go 5m/s at the top, but travels at 12m/min...

Last edited: Oct 19, 2007
7. Oct 19, 2007

### qspeechc

Your working out looks fine, and I expect the answer to be correct.
The key is that it is being lifted VERTICALLY at 12m/min. If the sand is being lifted at an agle to the horizontal (not equal to 90 degrees), then it will have a horizontal speed at the end, ie. the 5m/s. In fact, you can work out what this angle should be.