• Support PF! Buy your school textbooks, materials and every day products Here!

HELP: Work Done

  • Thread starter moimoi24
  • Start date
  • #1
48
0

Homework Statement


A box of weight 200N is pushed so that it moves at a steady speed alog a ramp, through a height of 1.5m. The ramp makes an angle of 30° with the ground. The frictional force on the box is 150N while the box is moving. What is the work doe by the person?


Homework Equations


none


The Attempt at a Solution


please explain this one.. thanks...
 

Answers and Replies

  • #2
CAF123
Gold Member
2,902
88
What have you tried? The box moves at steady speed so its change in kinetic energy is zero. There are two forces acting on the box which tend to transfer energy away from the box. THe person must put this energy in to allow the box to continue to move with steady speed.
 
  • #3
54
0
Lets solve this one

now there is no acceleration in the block right??

So as velocity is constant, there is no force

So the forces must cancel out...

hope now u get it
 
  • #4
48
0
i didnt get what you mean, sorry....

the answer is 750 J, but i am confused with the formula...
 
  • #5
CAF123
Gold Member
2,902
88
i didnt get what you mean, sorry....

the answer is 750 J, but i am confused with the formula...
Ok, you have identified the two forces transferring energy away from the block, right?
Compute the work done by those forces. (you should get negative values).
Therefore, if x amount of joules is transferred away then the person must provide that same number of joules into the system.
 
  • #6
48
0
this is how i work on it:

Total Force
F = Wsin30 + fr
= 200 (1/2) + 150
= 250 N

then, the path distance is 3m

W = F(s)
= 250 (3)
= 750 J
 
  • #7
48
0
Im just confused with the sin30 thing, why is it that formula? thanks a lot.
 
  • #8
54
0
Lemme try

now Frictional force =150N acting along the incline
and weight of the box acting along the incline = 200 *sin(30)=200*1/2=100N(weight*sin 30 gives us its weight along the incline and weight*cos 30 gives us weight perpendicular to incline)
So total force = 250 N along incline

so the man is also exerting a force of 250 N

now sin(30)=1.5/length of incline

so length of incline = 1.5 *2 =3m

so work done = 250*3 =750joules

Hope u got it now
 
  • #9
48
0
Lemme try

now Frictional force =150N acting along the incline
and weight of the box acting along the incline = 200 *sin(30)=200*1/2=100N(weight*sin 30 gives us its weight along the incline and weight*cos 30 gives us weight perpendicular to incline)
So total force = 250 N along incline

so the man is also exerting a force of 250 N

now sin(30)=1.5/length of incline

so length of incline = 1.5 *2 =3m

so work done = 250*3 =750joules




Hope u got it now

thanks... i mean, why do we have to use sin30?
 
  • #10
54
0
Do u know any program that can be used to make drawings ??
 
  • #11
48
0
what kind of drawing? autocad, paint, corel draw, photoshop....? please accept my friend invitation.. thanks...
 
  • #12
1,065
10
Total work done=Conservation force x height + Non-conservative force x d.
W=mgh + Ffd
W=200x1.5 + 150 x 1.5/Sin30
W=300 + 450
W=750 J.
 
  • #13
54
0
http://i1168.photobucket.com/albums/r493/siddhant35/th_b1ef28a8.jpg

now i can explain(sry for that messy thing)

if u zoom u can see that mg is acting downward and can be resolved into 2 components

1 along the incline and other perpendicular to it...(indicated by red)

after resolving u can get mg in 2 parts

here mgsin30 is acting along incline and mg cos 30 is acting along the perpendicular

note that mgcos 30 is cancelled out by normal force exerted by block

so mg sin 30 remains

now friction also acts along incline

so add forces along incline to get the force applied by man

then proceed as said above
 
Last edited:
  • #14
48
0
sorry but i cant see the image..
 
  • #15
54
0
Download and zoom silly
 
  • #16
48
0
thanks a lot... its a great help... :)
 
  • #17
54
0
some other tips for such problems

note that if friction is not given but coefficient of friction is given

as u know frictional force = coefficient of friction*Normal force

Here normal force would be mg cos 30

But this is not need here as friction is already given

hope that helped

hoping to see u around....
 
  • #18
48
0
please add me as your friend because i know i will still have lots of questions,,, thans for the help...
 

Related Threads on HELP: Work Done

  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
3
Views
562
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
5K
Replies
1
Views
742
  • Last Post
Replies
10
Views
13K
  • Last Post
Replies
10
Views
2K
Top