Question

Force and work.

Answer 1

A car drives at 800ft. on a road inclined at 9 degrees. Car weighs 2700lb. Thus gravity acts straight down on the car with a constant force of f=-2700j. Find the work done by the car in overcoming gravity.

Answer 2

Um..what i've done so far is
|v|=2700cos(9) = 2666.76
which is the force experienced by the driveway. Is that the answer i'm looking for?

Answer 3

@ganeshie8

Answer 4

@zepdrix :)

Answer 5

@jim_thompson5910

Answer 6

Work = Force * Displacement

Answer 7

Gravity is pulling straight down, so when we think of "displacement" we are only considering the vertical displacement of the car

Answer 8

when you wrote `Thus gravity acts straight down on the car with a constant force of f=-2700j`

that doesn't make any sense. Joules aren't a measure of force. They are measure of energy. You might be thinking of newtons?

Answer 9

uhh idk that's what my homework says.

Answer 10

oh nvm, -2700j is a vector, not 2700 joules

Answer 11

haha yeah

Answer 12

where is hte 800 from?

Answer 13

"A car drives at 800ft"

Answer 14

I'm assuming it drives up that incline and drives 800 ft

Answer 15

ah yeah k.

Answer 16

use trig to find the value of y in the drawing below

Answer 17

125.148 ?

Answer 18

so the car essentially goes up vertically 125.148 ft

this is the vertical displacement we care about

Answer 19

(ps: it says you are expected to use vectors and the appropriate formulas to find work. Just as a note. :P)

Answer 20

oo ok

Answer 21

so now we have
2=-2700 * 125.148

Answer 22

that 2 was meant to be a w

Answer 23

er. that will make a very nasty negative number..

Answer 24

were you given the vector formula for work in physics?

Answer 25

This is precalc.

Answer 26

give me a moment to find it, I have a formula somewhere :)

Answer 27

Yep
w=f*d

Answer 28

I'm thinking you basically take the force needed to climb that vertical distance and multiply it with the vertical distance

I'm not sure how to fit in vector formulas though

Answer 29

so an example given is:
A man pulls a wagon horizontally by exerting a force of 20 lb on the handle. If the handle makes an angle of 60 degrees with horizontal, find the work done in moving the wagon 100ft

Solution:
We choose a coordinate system with the origin at the initial position of the wagon. That is, the wagon moves from point p(0,0) to the point Q(100,0) the vector that represents this displacement is
d=100i

The force on the handle can be writen in terms of components as
F= (20cos(60))i + (20sin(60))j
F = 10i+10 sq3j
Thus the work done is

(10i+10 sq3j)*(100i)=1000

= 1000

Answer 30

so our displacement is
D = 800i

Answer 31

so it's written as
F=(2700cos9)i+(2700sin9)j
= 2666.785i+422.373j

Answer 32

and work done is =
(2666.785i+422.373j)*(800i)

Answer 33

I'm not sure how they calculated the last part though.

Answer 34

Is this kind of making sense? I'm following formulas mostly. o.0

Answer 35

@jim_thompson5910 :P

Answer 36

(sorry, i'll be patient ;))

Answer 37

sorry I got distracted for a sec

but it looks like you're on the right track. You dot product the force vector F and the displacement vector d to get the work

Answer 38

it's ok! I know you're helping other people. Keep up the hard work :)

Answer 39

so is
2133829.189 the final answer you're getting?

Answer 40

The force of gravity is

F = 0i + (-2700j)

essentially this vector pulls the car straight down with magnitude 2700 lb

--------------------

The displacement vector d = 800*cos(9)i + 800*sin(9)*j

--------------------

dot product the two vectors

W = F dot d

W = [0i + (-2700j)] dot [800*cos(9)i + 800*sin(9)*j]

W = [0*800*cos(9)] + [-2700*800*sin(9)]

W = -337,898.444486899

That seems like a pretty massive number, so I'm not 100% sure

Answer 41

bleh yeah that's pretty huge :/

Answer 42

wait why 800 instead of 2700 before the cos and sin?

Answer 43

because if we ignore the weight, and we just consider moving along the hypotenuse, we basically have this triangle

Answer 44

but in the example one they used the weight of the cart in that spot of the equation, which is why I ask.
hm but yeah I see what you're saying.