A force of F = 125 N is used to drag a crate 3.0 m across a floor. The force is directed at an angle upward from the crate so the vertical part of the force is Fv = 65 N and the horizontal part of the force is Fh = 107 N. How much work is done on the box?

Explanation:

The forward force to drag is 125N

Distance of 3m.

The distance is on the positive horizontal 3i+0j

The force is directed upward so that the horizontal and vertical component is

Fx=107N and Fy=65N

F=107i+65j

Work done is dot product of Force and distance

Then,

W=F.ds

W=(107i+65j).(3i+0j)

Note i.i=j.j=1, i.j=j.i=0

Therefore

W=107×3+65×0

W=321J

The workdone done on the box is 321J

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

c) power lifting, will be your answer.

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deku ❤