If an object is moving at $15 \frac{m}{s}$ $15 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{225}{g}$ $u_k=225 /g$ , how far will the object continue to move?

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Answer 1 · 2017-03-22T10:40:28

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

Kinetic friction is

$F_{k} = μ_{k} \cdot N$ $F_k=mu_k*N$

Normal reaction $= N g$ $=Ng$

$N = m g$ $N=mg$

So,

$F_{k} = μ_{k} \cdot m g$ $F_k=mu_k*mg$

$F = m a$ $F=ma$

$- F_{k} = - μ_{k} m g = m a$ $-F_k=-mu_kmg=ma$

$a = - μ_{k} g$ $a=-mu_kg$

$a = - \frac{225}{g} \cdot g = - 225 m s^{2}$ $a=-225/g*g=-225ms^2$

The negative sign indicates that this is a deceleration

Now, we apply the equation of motion

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

$v = 0$ $v=0$

$2 a s = - u^{2} = - 15^{2} = - 225$ $2as=-u^2=-15^2=-225$

$s = - \frac{225}{- 225} \cdot \frac{1}{2} = 0.5 m$ $s=-225/-225*1/2=0.5m$

If an object is moving at 15ms15 m/s over a surface with a kinetic friction coefficient of uk=225gu_k=225 /g, how far will the object continue to move?