Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I need help and fast

  1. Nov 16, 2006 #1
    hi i'm new here i wanted your help in solving these two questions i tried several times to solve them but the computer kept showing that the amswer is wrong ((on mastering physics site)) so please i need the answers to those problems and fast because the chapter will be closed in an hour.

    1.A factory worker pushes a 30.4kg crate a distance of 4.00 m along a level floor at constant velocity by pushing downward at an angle of 28.0 below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.255 .

    A.What magnitude of force must the worker apply to move the crate at constant velocity?
    Take the free fall acceleration to be 9.8
    B.How much work is done on the crate by this force when the crate is pushed a distance of 4.00 ?
    Take the free fall acceleration to be 9.8

    2.A pump is required to lift a mass of 790 of water per minute from a well of depth 13.8 and eject it with a speed of 18.1 .

    A.How much work is done per minute in lifting the water?
    Take the free fall acceleration to be = 9.80
    B.How much in giving the water the kinetic energy it has when ejected?
    C.What must be the power output of the pump?
    Take the free fall acceleration to be = 9.80 .

    NOTE; all the quantities are in SI units.
  2. jcsd
  3. Nov 16, 2006 #2
    please i tried hard i solved all my 5 problems but those two i just can't get them right
  4. Nov 16, 2006 #3
    show your work, lets see where you went wrong.
  5. Nov 16, 2006 #4
    ok the first one i tried to solve using this equation:
    Fcos(-28)-coe.k. (w) + Fsin(-28)=0

    and the answer was wrong so ofcourse i wasn't able to find the work

    the second problem was hard for me because i didn't take caculus at the first place
  6. Nov 16, 2006 #5
    why do you add Fsin(-28)?

    hint: pushing a box downwards and foreward on a plane when you don't have friction is like pushing it forward... (with only the horizontal component).
    when you got friction, and you push something downwards you add you force to its weight...
    Last edited: Nov 16, 2006
  7. Nov 16, 2006 #6


    User Avatar

    Staff: Mentor

    I don't understand your equation. Start with the fundamental equation

    [tex]F = \mu N[/tex]

    and express N as the sum of the weight of the crate plus the vertical component of the pushing force....
  8. Nov 17, 2006 #7
    i did that but it kept telling me it's wrong please help me i don't have much time it's due to 6 pm today:cry:
  9. Nov 17, 2006 #8
    You didn't do that in your earlier equation.

    Friction = uN, what is N? It is the weight of the crate + the additional force supplied by the worker. The weight of the crate is mg, and the additional force is the vertical component of his push, it's obvious you understand this, but you are not equating it to N, perhaps because you didn't draw a free body diagram?

    So, N = mg + Fv (Fv is the vertical part of the push, you will have to supply this), then

    Friction = u(mg + Fv)

    I agree with the other poster that you are making a mistake using -28 as the angle measure, just because it is below the horizontal doesn't make it a -28. It won't bother your cosine, but it will affect your sine function.

    I don't see any units in the second problem, sorry to say. How much water in how much time? Cubic meters? In any case, I don't think you need calculus for this.

    It's late. I'm going to bed.

  10. Nov 17, 2006 #9
    Hope you got it done neji006 ;)
  11. Nov 17, 2006 #10
    i did , boy i was nervous thanks guys i got 94% on the work and kinetic energy chapter.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook