If an object is moving at #48 m/s# over a surface with a kinetic friction coefficient of #u_k=3 /g#, how far will the object continue to move?

Question

If an object is moving at #48 m/s# over a surface with a kinetic friction coefficient of #u_k=3   /g#, how far will the object continue to move?

Leo Abeyta · Accepted Answer

It will continue to move about #384 m# long.

In this case, kinetic friction is an opposing force directed opposite to the direction of motion. For the above statement to be true, the acceleration acting on the object should be opposite (negative). Forming an equation for frictional force, #Nmu_k = Mg * 3/g = 3M# where, #N# = normal reaction = #Mg# #M# = mass of the object #:.# frictional force = #3M# According to my second statement, #3M = - Ma# So, #a = -3#m/#s^2# Now, substituting value of a in, #v = u + at# for an object to come at rest, its final velocity(v) must be = 0 So, #0 = 48 -3t# #t = 16s# Substituting the value of #v, u and a# in, #v^2 - u^2 = 2as# We get #s = 384m# Hence, the object travels #384m# long for #16s# Hope this helps !!!!

Olivia Askew · Answer

undefined To calculate the distance the object will continue to move, you can use the formula for kinetic friction force: $F_k = \mu_k \times N$, where $F_k$ is the kinetic friction force, $\mu_k$ is the coefficient of kinetic friction, and $N$ is the normal force. Then, use the equation of motion: $F_k = ma$ to find the acceleration $a$. Once you have the acceleration, you can use the equation of motion $v^2 = u^2 + 2as$ to find the distance $s$, where $v$ is the final velocity (0 m/s), $u$ is the initial velocity (48 m/s), and $a$ is the acceleration.