 #1
TheBigDig
 65
 2
 Homework Statement:

A person in a wheelchair (a total mass of 100 kg) is pushed from rest to a
speed of 2.0 m/s . If the effective coefficient of friction is 0.05 and the
distance moved is 15 m, what is the total work done?
 Relevant Equations:

##v^2 = u^2 +2as##
## F = ma ##
##f = \mu_k N##
##W = Fs##
Taking v = 2m/s, u=0m/s and s = 15m, we get
##a=0.13m/s^2##
##F_g = mg = 100(9.8) = 980N##
Since there's no vertical acceleration, the normal force is equal to the weight
##N = 980N##
##f = \mu_k N = 0.05(980) = 49N##
##F_{net} = ma = 100(0.13) = 13N##
##F_{app} = F_{net}+f = 62N##
My question is, is my work done the applied force by the distance or the net force by the distance. I'm a little unsure of it.