A constant 75-N force acts on a 6.00 kg box. The box initially is moving at 20 m/s in the direction of the force, and 3.0 s seconds later the box is moving at 58 m/s. Determine both the work done by this force and the average power delievered by the force during the 3.0-second interval.

Work done  = 8752 J

Power = 2917 W

Explanation:

Mass of the box, m= 6.00 kg

Force , F = 75 N

Initial velocity , vi = 20 m/s

The velocity after t = 3.0 s is vf = 58 m/s

Acceleration of the box is ,

a = (58m/s-20m/s)/3s

a = 12.7 m/s²

The distance moved by the box is

v² - vi² = 2as

(58)² -(20)^2 = 2*12.7 m/s^2 * s

s = 116.69 m

Work done, W = F.s

                     = 75 N * 116.69 m

                     = 8752 J

Power = W/t

           = 8752 J/3s

P = 2917 W

answer: he got 23

explanation:

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.