If an object with a mass of $10 k g$ $10 kg$ is moving on a surface at $15 \frac{m}{s}$ $15 m/s$ and slows to a halt after $4 s$ $4 s$ , what is the friction coefficient of the surface?

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Answer 1 · 2017-04-27T09:37:34

The coefficient of friction is $= 0.38$ $=0.38$

We start by calculating the deceleration.

We apply the equation of motion

$v = u + a t$ $v=u+at$

The initial velocity is $u = 15 m s^{- 1}$ $u=15ms^-1$

The time is $t = 4 s$ $t=4s$

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

Therefore,

$0 = 15 + 4 a$ $0=15+4a$

$a = - \frac{15}{4} m s^{- 2}$ $a=-15/4ms^-2$

According to Newton's Second Law, the frictional force is $F_{r} = m a$ $F_r=ma$

So,

$F_{r} = 10 \cdot \frac{15}{4} = \frac{75}{2} N$ $F_r=10*15/4=75/2N$

The normal force is

$N = m g = 10 \cdot 9.8 = 98 N$ $N=mg=10*9.8=98N$

The coefficient of friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{37.5}{98} = 0.38$ $mu_k=F_r/N=37.5/98=0.38$

If an object with a mass of 10 kg 10kg10 kg is moving on a surface at 15 m/s15ms15 m/s and slows to a halt after 4 s4s 4 s, what is the friction coefficient of the surface?