Find the work done by the force F 15i 6j newtons in moving an object 8 meters no

Question

Find the work done by the force F 15i 6j newtons in moving an object 8 meters north (i.e, in the j direction). The amount of work done is joules The vector w ai t bj is perpendicular to the line ax +by c and parallel to the line bx ay c. It is also true that the acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors that are either normal to the lines or parallel to the lines. Use this information to find the acute angle between the lines below. The angle is radians. (Type an exact answer, using as needed.)

Explanation / Answer

Since, I don't have a camera so I am explaining it to you manually:)

1)

work done = integral(F.X)

Since, force applied is constant all we have to do is calculate the dot product of force and displacement vector.

Given F=15i + 6j.

X=8j

so work done = F.X

=>work done = (15*0 +6*8)joules.

=>work done= 48 joules.

2)

Given two lines are:-

5x +9y=0.

7x + 2y =5.

now, equation of line parallel to line 2: -

7x +2y=0.

now take the slope of both lines:-

slope of line 1:-

m1= -5/9.

slope of line 2:-

m2=-7/2

let T be the acute angle between the two lines.

now acute angle T between line 1 and line 2 is given by the following formula:-

tan(T) = modulus[(m1-m2)/1+m1*m2].

---->tan(T)= 4.88

-->T= 1.368 radian

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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