If an object is moving at #2 m/s# over a surface with a kinetic friction coefficient of #u_k=6 /g#, how far will the object continue to move?

1 Answer
Feb 26, 2016

Answer:

#x=1/3m#

Explanation:

We can use the Energy Work Theorem to solve this equation.
#K_2-K_1=\int\vec{F}\cdot\hat{dx}#

Since the force and the distance travelled are in opposite directions, then #\vec{F}\cdot\hat{dx}=-Fdx#
Also, since it force works uniformly over the entire movement until rest of object (i,e till #K_2=0#), the integral will equate as such
#-int Fdx=-Fx#

So, all in all, it reduces to #K_1=Fx#

Now, force acting on the object is #F=\mu_kmg# (assuming it is a proper horizontal planar surface )
So, #1/2cancel{m}v_1^2=\mu_kcancel{m}gx\implies1/2v^2=\mu_kgx#
So, that means #x=v^2/(2\mu_kg#

Given #\mu_k=6/g\implies\mu_kg=6#, #v=2ms^-1\impliesv^2=4(ms^-1)^2#

Solving this gives the given answer above.