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    Q) Atractor of mass 500 kg pulls a trailer of mass 200 Kg up a rough slope inclined at 17 degrees to the horizontal. The resistance to the motion is 4N per kg. Calculate the work done by the tractor engine, given that the vehicle travels at a constant speed of 1.4m/s for 2 minutes.

    what i’ve done so far is calculate the distance travelled, s=(1.4)(120)=168 m

    then resolve the weight(in the direction of motion), 7000sin17

    then i calculated the resistance to motion, 4x700=2800

    then according to the work energy theorem, W(tractor)-7000sin17 (168) -2800 (168)=0

    but i just don’t seem to be getting the answer. Where have i gone wrong?
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    This looks right to me. Is your calculator in degree mode? What's the required answer?
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    I believe the resistance to motion bit includes gravity down the slope, so you dont have to add it manually.
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    (Original post by chailatte2407)
    This looks right to me. Is your calculator in degree mode? What's the required answer?
    my calculator is in degrees mode. The required answer is 814kJ but i keep getting 500something kJ
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    Wait no nvm. Is the answer 814kW?
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    (Original post by 1234123456785678)
    my calculator is in degrees mode. The required answer is 814kJ but i keep getting 500something kJ
    That's the answer I got using the same method as you. Check you're putting in the numbers correctly yo!
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    (Original post by Radioactivedecay)
    Wait no nvm. Is the answer 814kW?
    yes, how’d you get it?
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    I see where you went wrong. You dont take other forces into account when calculating the work done by a single force. this is because the work done by the force will be the same regardless of what other forces act opposite to it. for instance if you have a car that does x amount of work in One Direction that amount won't change if you have friction or something else pulling it back. therefore you only use the other forces to find out the force of the car that's pushing it and because it's a constant speed forces down the slope must equal forces up the slope. Find the power delivered by the car's engine using p=fv and then multiply that by the total time
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    (Original post by chailatte2407)
    That's the answer I got using the same method as you. Check you're putting in the numbers correctly yo!
    i think something’s wrong with my calculator then!
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    (Original post by 1234123456785678)
    i think something’s wrong with my calculator then!
    Look what you wrote:

    W(tractor)-7000sin17 (168) -2800 (168)=0

    Sooooo...

    W(tractor) = (7000sin(17) + 2800)*168

    Put that into a calculator and you get 814kJ!
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    (Original post by Radioactivedecay)
    I see where you went wrong. You dont take other forces into account when calculating the work done by a single force. this is because the work done by the force will be the same regardless of what other forces act opposite to it. for instance if you have a car that does x amount of work in One Direction that amount won't change if you have friction or something else pulling it back. therefore you only use the other forces to find out the force of the car that's pushing it and because it's a constant speed forces down the slope must equal forces up the slope. Find the power delivered by the car's engine using p=fv and then multiply that by the total time
    ahh that makes sense too! The thing is though i have to solve this using energy not power. I appreciate the help lots though!
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    (Original post by 1234123456785678)
    ahh that makes sense too! The thing is though i have to solve this using energy not power. I appreciate the help lots though!
    Power is energy per unit time ma dude, so same thing. You wont lose marks dw
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    (Original post by chailatte2407)
    Look what you wrote:

    W(tractor)-7000sin17 (168) -2800 (168)=0

    Sooooo...

    W(tractor) = (7000sin(17) + 2800)*168

    Put that into a calculator and you get 814kJ!
    I got that now!! i think i probably had an issue with the signs :/ Thanks a lot!!
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    (Original post by Radioactivedecay)
    Power is energy per unit time ma dude, so same thing. You wont lose marks dw
    it’s not that, i’ve just not gotten into power yet. (not a pastpaper, an exercise in the book in the section before power) I now could solve it using power but yeah i’d better know both ways
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