A12-pack of omni-cola (mass 4.30 kg) is initially at rest ona horizontal floor. it is then pushed in a straight line for1.20 m by a trained dog who exerts ahorizontal force with magnitude 39.0n.
(a) use the work-energy theorem to find the final speedof the 12-pack if there is no friction between the 12-pack and thefloor.
m/s
(b) use the work-energy theorem to find the final speed of the12-pack if the coefficient of kinetic friction between the 12-packand the floor is 0.30.
m/s

a)  W = 46.8 J  and b)   v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

    W = ΔK = -K₀

a) work is the scalar product of force by distance

    W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

    W = F d cos θ

    W = 39.0 1.20 cos 0

    W = 46.8 J

b) zero initial kinetic language because the package is stopped

    W - = -K₀

    W - fr d= ½ m v² - 0

    W - μ N d = ½ m v

   on the horizontal surface using Newton's second law

     N-W = 0

     N = W = mg

     W - μ mg d = ½ m v

    v² = (W -μ mg d) 2/m  

    v = √(W -μ mg d) 2/m

    v = √[(46.8 -  0.30 4.30 9.8 1.20) 2/4.3 ]

    v = √(31.63 2/4.3)

    v = 3.84 m/s

because if the tool that you are using is not balanced it there would be false mass.

one "light year" is the distance that light travels, through vacuum, in one julian year.   that's 365.25 days, of 24 hours each.