Question

An object is dragged 5m along the ground by a 47 N force acting at 50 degrees to the ground. It is then dragged 8 m up a ramp inclined at 30 degrees by the same exact force. Find the total work done.

Answer 1

wouldn't it be \[Fd _{1} \cos \theta _{1}+Fd _{2}\cos \theta _{2}\]

Answer 2

where F is force and d is distance in meters

Answer 3

yeah but I'd like to know how exactly I should draw the diagram

Answer 4

Use this formula...
Work=Force*distance
Where work is expressed in joules, force in newtons, and distance in meters.

Answer 5

idk that's a good question because are you still exerting the forrce at a 50 degree incline while on the 30 degree incline on the ramp?

Answer 6

I'm trying to help you solve it while in the midst of learning it too lol I'm in Physics right now as well

Answer 7

Ah tbh I really don't know since it doesn't say so in the question unfortunately :( it just asks for how much work is done in TOTAL

Answer 8

hmmmm. idk how many people are good with physics in this section but there's gotta b someone

Answer 9

Haha oops I thought It was a physics question. Srry @TheSweetLegend

Answer 10

Isn't it??

Answer 11

@ZeHanz

Answer 12

it's ok @e.cociuba! :) actually that was a formula my vectors teacher recommended the class using but I was curious as to why the angles were given.

Answer 13

it is

Answer 14

apparently this is what happens when you decide to not take physics :(

Answer 15

First part: \(W=F \cdot s=47\cos50⁰\cdot5\approx151.06~ J\)

Answer 16

Ohh ok I see.. lol @TheSweetLegend

Answer 17

Thanks @ZeHanz! :D so the 2nd part I do the same thing except for the ramp? and then I add them up to get the final answer right?

Answer 18

Yes, although I've not worked out yet how to calculate that second part.
There is a distance covered of 8 m, by a force of 47cos(20⁰).
But: the object is also moved up by 4 m against gravity, so I'd say that also involves some work...

Answer 19

oooohhh... so how should I solve that part?

Answer 20

and by 20 you mean 30 degrees from the ramp right?

Answer 21

the 20 degrees is with respect to the ramp (50-30=20).
Maybe it is better to calculate the work done by the horizontal movement and by the vertical movement separately.

Horizontal: \(4\sqrt{3}~m\cdot 47\cos50⁰~N \approx209.31~J\)
Vertical \(4 \cdot47\sin50⁰~N \approx 144.02~J
\)

Here, I have basically ignored the ramp and just looked at the point where the object arrives.

Now I'm not sure if I can just add the 209.31 and 144.02, or that I have to use the pythagorean theorem...