How much work would it take to horizontally accelerate an object with a mass of #3 kg# to #7 m/s# on a surface with a kinetic friction coefficient of #3 #?

1 Answer
Dec 28, 2017

Answer:

#mumgDeltax+1/2mv^2#

Explanation:

Use work-energy theorem.

#W_"net" = DeltaK#

If you ask how much 'net' work is done, then the answer is

#W_"net" = 1/2mv^2 - 1/2mv_0^2 =1/2mv^2 - 0= 1/2mv^2#

However, if you ask how much work is needed to overcome the friction to achieve #v=7m/s#, then you need to consider the net force and the distance traveled (#Deltax#):

#W_"net"= F_"net" Deltax = (F_"applied force" - F_mu)Deltax =DeltaK #

#F_mu = muN= muMg # is the frictional force

Therefore,

#F_"applied force"Deltax = F_muDeltax+ DeltaK= mumg Deltax + 1/2mv^2 #
The workdone by the applied force therefore does two things: to overcome the frictional work and to change the kinetic energy of the object.

However, neither x nor time t to reach #v=7m/s# are given. You need either one to solve this problem unless you meant net work done.