A mouse of mass 15g is stationary 2m below its hole. It runs to its hole, arriving 1.5 seconds later with a speed of 3ms^-1.

i) Show that the acceleration of the mouse it not constant. [Done by using s=ut+.5at^2 and v=u+at and showing that they give different answers for acceleration]

ii) Calculate the average power of the mouse

Since power = work done / time and work done = .5mv^2 - .5mu^2, power = ((0.5*0.015*3^2) - 0) / 1.5

=0.045W.

However, the book gives the answer 0.241W.