- #1

Julie pulls a 61.5 N bag a distance of 278 m at a constant velocity. The force she exerts is 48.4 N at an angle of 66.8 degrees above the horizontal.

I found the work julie does on the bag to be 5300 J.

Yet I am not sure how to find A) the work done by the force of friction on the bag and B) the coefficient of kinetic friction on the bag

Next problem is

A rain cloud contains 2.82 x 10^7 kg of water vapor.

The acceleration of gravity is 9.81m/s^2

How long would it take a 2.10 kW pump to raise the same amount of water to the cloud's altitude of 1.44km? Answer in seconds (s)

[I tried Power= work(FsCOS[theta]) divided by t. So 2100w=2.82x10^7(9.81)1440m COS 90 degrees DIVIDED by t, which just equals 0 since cos90 is 0. I DONT GET IT!]