A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction between the box and the floor is 0.20. The box moves a distance of 4.0 m in 2.0 s. The magnitude of the change in momentum of the box during this time is most nearly .

Explanation:

The given data is as follows.

       initial speed (u) = 4.0 m/s,     mass (m) = 12 kg

      Distance (s) = 4.0 m,    time (t) = 2.0 sec

First, we will calculate the acceleration as follows.

           s =

         4 =

         a = -2

Now, the final speed will be calculated as follows.

            v = u + at

               =

              = 0

Therefore, change in momentum will be calculated as follows.

            = m(v - u)

                           =

                           = -48 kg m/s

The negative sign indicates the change in momentum.

Thus, we can conclude that the change in momentum of the box during this time is most nearly 48 kg m/s.

d is the answer i came up with

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.