A 3-lb force acting in the direction of the vector (3,-1) moves an object just over 9 ft from point (0,5) to (6,-2). Find the work done to move the object to the nearest foot-pound
Accepted Solution
A:
Answer: The Work done to move the object is [tex]=23.15 \text { foot pounds }[/tex]Step-by-step explanation:Given;Force F=3 [tex]l b s[/tex]Vectors of F= (3,-1)Moves object to a distance=9 [tex]f t s[/tex]Let A and B be the displacements VectorsA= (0,-5)B= (6,-2)To Find:Work done in foot-poundsSolution:Work done [tex]W=F \times D[/tex]Direction of force vector= (3,-1) (i, j) =3i-jUnit of force vector [tex]=\sqrt{3^{2}+\left(-1^{2}\right)}[/tex] [tex]=\sqrt{9+1}[/tex]Force vector=(Force/Unit Vector)Direction of force vectors F=[tex]3 / \sqrt{10} \times(3 i-j)[/tex]Direction of motion vector= (B-A) = (6,-2)-(0,-5) x i,j) =(6-0),(-2-5) x (i, j) =(6,-7) x (i, j) =6i-7jUnit of motion vector [tex]=\sqrt{6^{2}+\left(-7^{2}\right)}[/tex] [tex]=\sqrt{36+49}[/tex] [tex]=\sqrt{85}[/tex]Motion Vector= (Distance moved by the object/Unit motion vector) × (Direction of motion vectors) [tex]D=9 / \sqrt{85} \times(6 i-7 j)[/tex]Workdone [tex]W=F \times D[/tex] [tex]= [3/ \sqrt{10} \times (3i-j)] \times [9/ \sqrt{85} \times (6i-7j)][/tex] [tex]= [3/ \sqrt{10} \times 9/ \sqrt{85}] \times [(3i-j) \times(6i-7j)][/tex] [tex]= [3/3.1623 \times 9/9.2195] \times [(3\times6) + ((-1) \times (-7))][/tex] [tex]= [0.94867 \times 0.97619] \times [18+7][/tex] [tex]=23.15 \text { foot pounds }[/tex]Result: Work done to move an object [tex]=23.15 \text { foot pounds }[/tex]