If a $8 k g$ $8 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows to a halt after moving $80 m$ $80 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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If a $8 k g$ $8 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows to a halt after moving $80 m$ $80 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Answer 1 · 2018-01-20T18:20:54

0.16

Explanation:

Clearly,here the frictional force caused retardation of the moving object and brought it to rest.

So,using $v^{2} = u^{2} - 2 a s$ $v^2 = u^2 - 2as$ (all symbols are bearing their conventional meaning)(given, $v = 0$ $v=0$ , $u = 16 \frac{m}{s}$ $u=16 m/s$ and $s = 80$ $s = 80$ )

We get, $a = 1.6 \frac{m}{s^{2}}$ $a = 1.6 m/s^2$

Now,if the coefficient of kinetic friction is $μ$ $mu$ , frictional force( $f$ $f$ ) acting on the body was $μ \cdot N$ $mu*N$ or $μ \cdot m g$ $mu*mg$ i.e $μ \cdot 80$ $mu*80$

So,we using $f = m a$ $f = ma$ (where, $m = 8 K g$ $m=8Kg$ and $a = 1.6 \frac{m}{s^{2}}$ $a = 1.6 m/s^2$ )

We, get $μ \cdot 80 = 12.8$ $mu*80 = 12.8$
Or, $μ = 0.16$ $mu = 0.16$

Answer 2 · 2018-01-20T18:23:37

The coefficient of kinetic friction is $= 0.16$ $=0.16$

Explanation:

The mass of the object is $m = 8 k g$ $m=8kg$

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

The Normal reaction is

$N = m g = 8 \times 9.8 = 78.4 N$ $N=mg=8xx9.8=78.4N$

The initial velocity is $u = 16 m s^{- 1}$ $u=16ms^-1$

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

The distance is $s = 80 m$ $s=80m$

Apply the equation of motion,

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

The acceleration is

$a = \frac{v^{2} - u^{2}}{2 s} = \frac{0^{2} - 16^{2}}{2 \cdot 80} = - 1.6 m s^{- 2}$ $a=(v^2-u^2)/(2s)=(0^2-16^2)/(2*80)=-1.6ms^-2$

According to Newton's Second Law

$F = m a$ $F=ma$

The force of friction is

$F_{r} = 8 \cdot 1.6 = 12.8 N$ $F_r=8*1.6=12.8N$

The coefficient of kinetic friction is

$μ_{k} = \frac{F_{r}}{N} = \frac{12.8}{78.4} = 0.16$ $mu_k=F_r/N=12.8/78.4=0.16$

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If a $8 k g$ $8 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows to a halt after moving $80 m$ $80 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?

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Explanation:

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Science

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If a 8kg8 kg object moving at 16ms16 m/s slows to a halt after moving 80m80 m, what is the coefficient of kinetic friction of the surface that the object was moving over?

2 Answers

Explanation:

Explanation:

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If a $8 k g$ $8 kg$ object moving at $16 \frac{m}{s}$ $16 m/s$ slows to a halt after moving $80 m$ $80 m$ , what is the coefficient of kinetic friction of the surface that the object was moving over?