A force F= (6.70 N)i + (5.60 N)j + (6.60 N)k act on a 9.80 kg mobile object that moves from an initial position of di = (9.10 m)i + (5.30 m)j + (6.70 m)k to a final position of df = (8.00 m)i + (9.70 m)j + ( 3.30 m)k in 5.90 s.

Find (a) the work done on the object by the force in the 5.90 s interval, (b) the average power due to the force during that interval.

W = 0.11J

P = 0.0186W

Explanation:

Please see attachment below.

v=ir = 3x15 =45 v

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards