Question

If an object is moving at `2 m/s` over a surface with a kinetic friction coefficient of `u_k=17 /g`, how far will the object continue to move?

Answer 1

`x=2/17=0.11764" "`meter

Explanation:

Let `x` be the distance

From `v_f^2=v_0^2+2ax`

`-F_k=m*a" "`Kinetic Frictional force
`a=(-F_k)/m" "` acceleration

`F_k=mu_k*F_n" "` Relation between Kinetic frictional force and the normal force

`F_n=m*g`

`x=(v_f^2-v_0^2)/(2a)`

`x=(v_f^2-v_0^2)/(2((-F_k)/m))`

`x=(v_f^2-v_0^2)/(2((-(mu_k*F_n))/m))`

`x=(v_f^2-v_0^2)/(2((-(mu_k*m*g))/m))`

`x=(v_f^2-v_0^2)/(2(-mu_k*g))`

`v_f=0" "`when it stops
`v_0=2" "`meter /second
`mu_k=17/g`

`x=(v_f^2-v_0^2)/(2(-mu_k*g))`
`x=(0^2-2^2)/(2(-(17/g)*g))`

`x=(-4)/(-34)=2/17" "`meters

God bless....I hope the explanation is useful.