# Work Power and Energy Help

1. Apr 24, 2006

### kaity

Ok so I'm in Gr 11 U Physics and we have an assignment and I have a couple q's

1) While at the park a child sits on top of a 3.5m slide. When he reaches the bottom of the slide, he is travelling at 3.0m/s. Calaculate the efficency of the slide

2) A 5000 W motor is lifting crates from the bottom of a mine shaft to the ground above at a constant rate of 5m/s. What is the maximum weight of a crate to be lifted up the shaft?

2. Apr 24, 2006

### Hootenanny

Staff Emeritus
Welcome to PF Kaity,

The policies of this forum prevent us from helping you, unless you show some working or initial thoughts...

~H

3. Apr 24, 2006

### kaity

ok well for the first one...we did something similar in class, but when I tried it, it was definetly wrong. we were told to not include mass when it is not given so for this it would be:
Eg = mg x h Ek= 1/2mv2
= m(9.8)(3.5) =(0.5)m(3.0)2
=34.3mJ =4.5mJ

%e =Eg /Ek x 100%
= 34.3mJ/ 4.5m/J x100%
= 762.2%This answer is evidently wrong

4. Apr 24, 2006

### kaity

for the second one I have no idea where to even begin
P=5000W or J/s
v= 5.0m/s

So I originally tried figuring out force which you can get by doing
F = P/ V
= 5000W/5.0m/s
= 1000N
and I believe you can find the mass of that by dividing the
1000N/9.8 = 102.04kg
is that it or is there more to it?

5. Apr 25, 2006

### Hootenanny

Staff Emeritus
First, you need to find the energy lost;

$$E_{lost} = E_{inital} - E_{final$$

Then you need to find the efficency;

$$= 1- \frac{E_{lost}}{E_{initial}}$$

Can you go from here?

~H

6. Apr 25, 2006

### illwerral

First off, I solved the 5000W motor problem using different equations and arrived at the same answer, so I would say that your calculation of the mass is correct. What I did was realize that since power is work over time, and the work will be the increase in potential energy in this case I could replace work with mgh (where m is mass, g is acceleration due to gravity and h is the change in height) and then solve for m. That method yielded your answer.

For the slide question, you stated that %e=Eg/Ek*100. In my class we said that %e=(energy out)/(energy in) * 100. I think that will give you a much more reasonable result.