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?

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?

Isabella Alvarez · Accepted Answer

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

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.

Axel Acevedo · Answer

undefined The distance the object will continue to move can be calculated using the formula: $ d = \frac{{v^2}}{{2 \cdot u_k \cdot g}} $, where $ d $ is the distance, $ v $ is the initial velocity, $ u_k $ is the kinetic friction coefficient, and $ g $ is the acceleration due to gravity. Plugging in the values, you get $ d = \frac{{2^2}}{{2 \cdot 0.17 \cdot 9.8}} $. Calculate to find the distance.