A constant force of 12 N in the +x direction acts on a 4 kg block as it moves from the origin to the point (6 − 8) m. How much work is done on the block by this force during this displacement?

72 Nm²

Explanation:

work = F . d

work = (12i ,0j) . (6i ,-8j )

work = 72 J

The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.

It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j

When a force vector F acts in a nonparallel direction d, the work is given by:

W = |F|*|d|*cos(theta) where theta is the angle between the vectors.

alternatively you can use the dot product of the two vectors to get:

W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.