If an object is moving at $1 \frac{m}{s}$ $1 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{1}{g}$ $u_k=1 /g$ , how far will the object continue to move?

Question

1391 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-31T07:01:32

The distance is $= 0.5 m$ $=0.5m$

The initial velocity is $u = 1 m s^{- 1}$ $u=1ms^-1$

The final velocity is $v = 0 m s^{- 1}$ $v=0ms^-1$

The distance is $s = ? m$ $s=?m$

The coefficient of friction is

$μ_{k} = \frac{F_{r}}{N}$ $mu_k=F_r/N$

Let the mass be $= m k g$ $=mkg$

Then

the normal reaction is $N = m g N$ $N=mgN$

So,

$F_{r} = μ_{k} \cdot m g N$ $F_r=mu_k*mgN$

$F = m a$ $F=ma$

Therefore,

$m a = - μ_{k} m g$ $ma=-mu_kmg$

The acceleration is $a = - μ_{k} g$ $a=-mu_kg$ , the negative sign indicates a deceleration

We apply the equation of motion

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$0 = 1^{2} - 2 \cdot \frac{1}{g} \cdot g \cdot s$ $0=1^2-2*1/g*g*s$

$s = \frac{1}{2} = 0.5 m$ $s=1/2=0.5m$

If an object is moving at 1ms1 m/s over a surface with a kinetic friction coefficient of uk=1gu_k=1 /g, how far will the object continue to move?