A $2 k g$ block sliding on horizontal surface with a speed of $5 m / s$ comes to rest after covering distance of $5 m$ . Assuming that the retardation is uniform find $μ_{2}$

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Solution

The equation of motion is given as,

$v^{2} - u^{2} = 2 a s$

$0 - {(5)}^{2} = 2 \times a \times 5$

$a = - 2.5 m / s^{2}$

When the block comes to rest, the force equation is given as,

$f + m a = 0$

$μ m g + m a = 0$

$μ \times 10 + (- 2.5) = 0$

$μ = 0.25$

Thus, the coefficient of friction is $0.25$ .

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A 2kg block sliding on horizontal surface with a speed of 5m/s comes to rest after covering distance of 5m . Assuming that the retardation is uniform find μ2

The equation of motion is given as, v2−u2=2as 0−(5)2=2×a×5 a=−2.5m/s2 When the block comes to rest, the force equation is given as, f+ma=0 μmg+ma=0 μ×10+(−2.5)=0 μ=0.25 Thus, the coefficient of friction is 0.25.

A $2 k g$ block sliding on horizontal surface with a speed of $5 m / s$ comes to rest after covering distance of $5 m$ . Assuming that the retardation is uniform find $μ_{2}$