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
Ashley H
Aug 11, 2018

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

Narad T.
Aug 11, 2018

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

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