1. Help!

You lift a 10. lb physics book up in the air a distance of 1 ft. at a constant velocity of 0.5 ft/s. the work done by gravity is?

2. Originally Posted by babygirl
You lift a 10. lb physics book up in the air a distance of 1 ft. at a constant velocity of 0.5 ft/s. the work done by gravity is?
First note that we don't need to know that the book is moving at a constant velocity.

The angle between the weight and displacement is 180 degrees.
$W_G = \vec w \cdot \vec s = (mg) \Delta h \, cos(180) = -mg \Delta \, h$

Thus the work done by gravity is = -10*1 ft-lbs = -10 ft-lbs.

-Dan

(And what Physics book is still using English units??)

3. Originally Posted by topsquark
First note that we don't need to know that the book is moving at a constant velocity.

The angle between the weight and displacement is 180 degrees.
$W_G = \vec w \cdot \vec s = (mg) \Delta h \, cos(180) = -mg \Delta \, h$

Thus the work done by gravity is = -10*1 ft-lbs = -10 ft-lbs.
I think that an explanation of why $g$ is $1$ in this
unit system, and that using $g=32 ft/s^2$ would give an answer
in $ft-poundals$ (or what ever the units are - its >30 years since
I used them).

-Dan

(And what Physics book is still using English units??)
Quite - Mars probes have been lost over this sort of thing (revenge of the
killer-slugs (kilo-slugs) I expect).

RonL

4. Originally Posted by CaptainBlack
Quite - Mars probes have been lost over this sort of thing (revenge of the
killer-slugs (kilo-slugs) I expect).
Heard that story a long time ago from my father. So funny, how embarrasing is that

5. Originally Posted by CaptainBlack
I think that an explanation of why $g$ is $1$ in this
unit system, and that using $g=32 ft/s^2$ would give an answer
in $ft-poundals$ (or what ever the units are - its >30 years since
I used them).
Actually, g IS 32 ft/s^2. Note that the book's weight is given, not its mass. (ie. lb is a unit of weight, not mass). Thus mg = 10 lb, so the value of g is not directly used here. The 1 in my equation was the "1 ft." I suppose I should have labelled the units as I put them in the equation to make that clear.

I've honestly forgotten how a foot-poundal is defined. I'll have to look that up sometime.

-Dan

6. Originally Posted by topsquark
Actually, g IS 32 ft/s^2. Note that the book's weight is given, not its mass. (ie. lb is a unit of weight, not mass). Thus mg = 10 lb, so the value of g is not directly used here. The 1 in my equation was the "1 ft." I suppose I should have labelled the units as I put them in the equation to make that clear.

I've honestly forgotten how a foot-poundal is defined. I'll have to look that up sometime.

-Dan
Its a dreadful feature of the Customary/Imperial Unit system that the pound is
commonly used for weight, it is in that system's the unit of Mass. The unit of
force is the poundal - the force that would accelerate a MASS of 1 pound
at 1 ft/s^2.

RonL

7. Originally Posted by ThePerfectHacker
Heard that story a long time ago from my father. So funny, how embarrasing is that
Good grief - now I feel old

RonL

8. Originally Posted by CaptainBlack
Its a dreadful feature of the Customary/Imperial Unit system that the pound is
commonly used for weight, it is in that system's the unit of Mass. The unit of
force is the poundal - the force that would accelerate a MASS of 1 pound
at 1 ft/s^2.

RonL
Oh dear, I didn't know that. Okay, so we're both right!

-Dan

(I knew there was a reason we were switching over to metric!)

9. Originally Posted by topsquark
(I knew there was a reason we were switching over to metric!)
You know, even though the customary system is HORRIBLE for math, it is quite usefull in approximating distance in everyday life without using a ruler...

1 inch=the length of 1 digit in your finger (I think your thumb)
1 foot=the length of your foot
1 yard=the length of a long stride
1 mile=I have no idea (probably the distance from two ancient cities)