If an object is moving at #50 m/s# over a surface with a kinetic friction coefficient of #u_k=2 /g#, how far will the object continue to move?

2 Answers
Aug 11, 2018

Answer:

#625m#

Explanation:

Because of kinetic frictional force,the object will undergo constant deceleration.

So,to find upto what length it will go,we can apply the relation between velocity (#v#),deceleration (#a#) and displacement (#s#)

i.e #v^2=u^2-2as #(where, #u# is the initial velocity)

Now, frictional force acting is #f=mumg=2/g×mg=2m# where, #m# is its mass.

So, deceleration i.e #a=f/m=2ms^-2#

Putting in the equation and also putting #v=0# as it will travel until its final velocity becomes zero.

So,we get, #0^2=50^2-2×2×s#

Or, #s=625m#

Aug 11, 2018

Answer:

The distance is #=625m#

Explanation:

The coefficient of kinetic friction is

#mu_k=F_r/N#

The mass of the object is #=m#

The acceleration due to gravity is #g=9.8ms^-2#

The normal reaction is #N=mg#

Therefore,

#F_r=mu_kN=mu_kmg#

According to Newton's Second Law

#F=ma#

The acceleration is

#a=F/m=-F_r/m=-(mu_kmg)/m=-mu_kg#

The coefficient of kinetic friction is #mu_k=2/g#

#a=-2/g*g=-2ms^-2#

The initial velocity is #u=50ms^-1#

The final velocity is #v=0ms^-1#

Apply the equation of motion

#v^2=u^2+2as#

The distance is

#s=(v^2-u^2)/(2a)#

#=(0-50^2)/(2*-2)#

#=50^2/4#

#=625m#