If an object with a mass of $10 k g$ $10 kg$ is moving on a surface at $45 \frac{m}{s}$ $45 m/s$ and slows to a halt after $4 s$ $4 s$ , what is the friction coefficient of the surface?

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Answer 1 · 2018-06-06T05:49:20

The coefficient of friction is $= 1.15$ $=1.15$

Calculate the acceleration with the equation

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

The initial velocity is $u = 45 m s^{- 1}$ $u=45ms^-1$

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

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

The acceleration is

$a = \frac{v - u}{t} = \frac{0 - 45}{4} = - 11.25 m s^{- 2}$ $a=(v-u)/t=(0-45)/4=-11.25ms^-2$

The mass of the object is $m = 10 k g$ $m=10kg$

Apply Newton's Second Law to calculate the frictional force

$F_{r} = m \cdot | a | = 10 \cdot 11.25 = 112.5 N$ $F_r=m*|a|=10*11.25=112.5N$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The normal force (reaction) is $N = m g = 10 \cdot 9.8 = 98 N$ $N=mg=10*9.8=98N$

The coefficient of friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{112.5}{98} = 1.15$ $mu_k=F_r/N=112.5/98=1.15$

If an object with a mass of 10 kg 10kg10 kg is moving on a surface at 45 m/s45ms45 m/s and slows to a halt after 4 s4s 4 s, what is the friction coefficient of the surface?