A200 kg crate is pulled along a level surface by an engine. the coefficient of friction between crate and surface is 0.4.
a. how much power must the engine deliver to move the crate at 5.0m/s.
b. how much work is done by the engine in 180s

a)Power = P = 3924 watts

b)Work = W = 706320 J

Explanation:

a)Mass of the block = m = 200 kg

       Velocity = V = 5 m/s

       Time = t = 180s

       Coefficient of friction = µ = 0.4

       Gravitational acceleration = g = 9.81 m/s^2

        According to newton’s second law of motion:

                              F-f = 0

        Where,

         F = Normal force  

         f = Frictional force

          so,  

                          F = f = µmg = (0.4)(200)(9.81) =  784.8 N

          We know that:

                         P = FV = (784)(5) = 3924 watts

b)      We know that:

          W = P × t = 3924 × 180 = 706320 J