Energy, work and power question

Mar 2017
I can't work out why I'm getting the part ii wrong and I was wondering if you could help.

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

However, the book gives the answer 0.241W.


PHF Helper
Jun 2010
Morristown, NJ USA
Don't forget that the work done includes both the gain in KE and the gain in PE as the mouse runs up hill.
Mar 2017
Oh, of course! :)

power = ((0.5*0.015*3^2) + 0.015g*2) / 1.5 = 0.241W

Thanks ChipB