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Posted by on Monday, May 7, 2012 at 11:37pm.

Find the total work done by a 15 N force in the direction of the vector [1,2,2], when it moves a particle from O[0,0,0] to P[1,-3,4] and then from P to A[7,2,5]. Distance is measured in metres.

  • Calculus - , Tuesday, May 8, 2012 at 2:14pm

    Work done by a force F along a vector P is F.P (dot product). So total work done is
    F.P + F.A
    where A=<1,-3,4>, P=<7-1, 2-(-3), 5-4>
    and F=15<1,2,2>/sqrt(1^2+2^2+2^2)=<5,10,10>
    The division by sqrt(...) is required to normalize the direction vector to a unit vector.

  • Calculus - , Wednesday, May 9, 2012 at 1:11am

    I don't understand it.. what is the final work done? Also, the arrows (< and >) are they actually supposed to be written as < and > or are they supposed to mean brackets ( ) or these brackets []

    thank you

  • Calculus - , Wednesday, May 9, 2012 at 3:40pm

    Most books write vectors as <x,y,z> to differentiate from a point (x,y,z).
    You would separate the work done by P along the two lines (vectors) OP and PA.

    The force is 15N along <1,2,2>. Since <1,2,2> is not a unit vector, we need to normalize it as
    <1,2,2>/sqrt(1^2+2^2+2^2)=<1,2,2>/3
    So the force is
    F=15<1,2,2>/3 = 5<1,2,2>=<5,10,10>

    Work done is F.D, where F and D are vectors, and . represents the dot product.

    The first part of the work is
    F.OP=<5,10,10>.<1-0,-3-0,4-0>
    =<5,10,10>.<1,-3,4>
    =5*1-10*3+10*4
    =15

    I'll leave it to you to calculate the second part, F.PA, where
    A=<7-1, 2-(-3), 5-4>=<6,5,1>

  • Calculus - , Thursday, May 10, 2012 at 4:05pm

    could you help me calculate the second part also? thank you

  • Calculus - , Thursday, May 10, 2012 at 8:53pm

    F.PA=<5,10,10><6,5,1>
    =5*6+10*5+10*1
    =30+50+1
    =81

  • Calculus - , Saturday, May 19, 2012 at 9:43am

    Isnt it 30 + 50 + 10 because 10* 1 is 10 so final is = 90

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