A particle moves along a straight path through a displacement d = 1.5i + cj while a force F = -8.5i + -6.5j acts on it. The displacement is measured in meters and the force is measured in Newtons. (Other forces also act on the particle.)
1. What is the value of c if the work done by F on the particle is (a) zero, (b) 7.8 J, and (c) -5.6 J

(a) c = - 1.96

(b) c = - 3.16

(c) c = - 1.1

The work done (W) on a particle is the dot product of its force vector (F) and its displacement vector (d).

i.e  

W = F . d          (i)

Given in the question;

F = -8.5i + -6.5j

d = 1.5i + cj

Calculate the dot product of F and d as follows;

F . d =  (-8.5i + -6.5j) . (1.5i + cj)

F . d =  (-8.5 x 1.5) i² + (-6.5 x c) j²    Note:  [i² = j² = 1]

F . d =  -12.75 - 6.5c

Substitute F . d =  -12.75 - 6.5c  into equation (i) as follows;

W = -12.75 - 6.5c    (ii)

(a) when W = 0

Substitute W = 0 into equation (ii) as follows;

0 = -12.75 - 6.5c

6.5c = - 12.75  

Solve for c;

c = - 12.75 / 6.5

c = - 1.96

(b) when W = 7.8

Substitute W = 7.8 into equation (ii) as follows;

7.8 = -12.75 - 6.5c

6.5c = - 12.75 - 7.8

6.5c = - 20.55

Solve for c;

c = - 20.55 / 6.5

c = - 3.16

(c) when W = -5.6

Substitute W = -5.6 into equation (ii) as follows;

-5.6 = -12.75 - 6.5c

6.5c = - 12.75 + 5.6

6.5c = - 7.15

Solve for c;

c = - 7.15 / 6.5

c = - 1.1

answer; (arch); good luck

the total distance that martina travels that day is

(distance from home to mom's office) plus (distance from mom's office to school)

plus (distance from school to library) plus (distance from library to friend's house)

plus (distance from friend's house to martina's house).

(we don't know what any of these distance actually is.)