How much work Is done by a force F = (7 Times N)1 + (3 N)j, with x in meters, th

Question

How much work Is done by a force F = (7 Times N)1 + (3 N)j, with x in meters, that moves a particle from a position T_1 = (9 m) t + (3 m) j to position T_r = (-7 m)I + (-2 m)j? I am confused with the 4x force. Does anyone have advice for setting this up? I did not understand it either. I just Ignored it and got the right answer, use W-(Fx dx)+(Fy dy) I tried to ignore It too but I keep getting the wrong answer. You don't get the right answer when you ignore x, However, we do have the equation to make our lives easier.

Explanation / Answer

We can find the x component of work and the y and add them

WX

integral (F *dx) evaluated from 9 to -7

= int (7x*dx) = 7x2/2 = 3.5x2 = 3.5*((-7)2 - 92) = -112J

Wy

is given by 3*(-2-3) = -15J

so the work is -112 - 15 = -127J

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.