Question

Two Physics Questions! Please help!! 1. A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact southeast with a force of 2.00 × 104 N. The sailboat travels the distance in 1.0 h. How much work was done by the wind? What was the wind’s power? Your response should include all of your work and a free-body diagram.

2. A sailor pushes a 100.0 kg crate up a ramp that is 3.00 m high and 5.00 m long onto the deck of a ship. He exerts a 650.0 N force parallel to the ramp. What is the mechanical advantage of the ramp? What is the efficiency of the ramp? Your response sh

Answer 1

@petiteme @wolfe8

Answer 2

anyone?

Answer 3

@petiteme

Answer 4

Petite is asleep. So can you at least draw the diagram?

Answer 5

so far I have work = force x distance x cos theta?

Answer 6

\[W=F\Delta x\]
\[P=\frac W t\]

Answer 7

okay, yeah, if you have an angle then you need the cosine

Answer 8

Ok but what about the diagram with the forces and direction labeled?

Answer 9

the first thing you need to do is establish a coordinate system

Answer 10

oh and next time, please only post one question at a time
it gets confusing when you try answering two at once

Answer 11

so ok, you have two directions to consider, north and northwest
the wind is blowing FROM the southeast, relative to the boat so the wind is going north west

Answer 12

you need to establish a coordinate system in order to figure out which direction the boat is going and which direction the wind is blowing. once you have that, you can figure out the angle theta

Answer 13

ok

Answer 14

once you do that, you know that \[W=F d \cos\theta\]
you know how long the wind was blowing for and that
\[Power= \frac W t\]
so this is mostly a substitution problem

Answer 15

This might help http://sketchtoy.com/58720506 Now you need to resolve the force that actually helped act on the movement north so resolve into the y-component.

Answer 16

Sorry for the mistakes. I was trying to figure out what values you need.

Answer 17

its ok so force = 20,000
distance = 10.00 km

Answer 18

so work = 20,000 x 10.00 x cos theta which would be 135?

Answer 19

...how on earth did you get 20,000? o.o You do Fy=Fcos(45) = 208 cos (45)

Answer 20

That will give you the force. And them you multiply it with distance to get work. To get power, use work/time

Answer 21

I thought since it said force of 2.00 x 10^4

Answer 22

ohh my bad I think I wrote 2.00 x 104

Answer 23

...when you copy and pasted it it became 2x104 N Ok then you should do 20000 cos (45)

Answer 24

wait what about the distance?

Answer 25

distance is given

Answer 26

so wouldn't it be 20,000 x 10.0 x cos 45?

Answer 27

You till multiply the resolved force with distance to get work because W=Fd

Answer 28

I guess you can write it that way when combined. I was trying to show you step-by-step

Answer 29

@wolfe8 so it would be 141,421.36 J for work?

Answer 30

Looks right to me

Answer 31

ok so for power what would it be?

Answer 32

@petiteme @wolfe8

Answer 33

Sigh. I told you every step you need. For power it is W/t Work divided by time.

Answer 34

yes I know what work is now but time? just 1?

Answer 35

@wolfe8

Answer 36

...

Answer 37

Time in seconds so first you must convert 1 hour to seconds.

Answer 38

so 60 s

Answer 39

2,357.03 W?

Answer 40

or 3600s ?

Answer 41

sigh

Answer 42

In 1 hour there are 3600 seconds

Answer 43

ohh sorry I was thinking one minute

Answer 44

39.28 W?