# Find coefficient of kinetic friction?

## A 18 kg block is released from point A as shown in the figure.  Its path from A to the spring is frictionless except for the portion between B and C.  When the block hits the spring, it compresses the spring by 0.3 m from its equilibrium position before coming to rest momentarily.  Find the coefficient of kinetic friction between the block and the rough surface loacted between points B and C.  Assume the spring constant is 2500 N/m.

Jun 27, 2018

0.394

#### Explanation:

Applying conservation of energy from A to B;

$m g h = \frac{1}{2} m \cdot {u}^{2} \Rightarrow \cancel{m} g h = \frac{1}{2} \cancel{m} \cdot {u}^{2} \Rightarrow {u}^{2} = 2 g h$

In B to C,

${u}^{2} = 2 g h$

$a = - g {\mu}_{k}$

${v}^{2} = {u}^{2} + 2 a s \Rightarrow {v}^{2} = 2 g h - 2 s g {\mu}_{k} \Rightarrow {v}^{2} = 2 g \left(h - s {\mu}_{k}\right)$

Applying conservation of energy from C to rest

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2} \Rightarrow 2 m g \left(h - s {\mu}_{k}\right) = k {x}^{2}$

substituting $m = 18 k g , g = 9.8 m \cdot {s}^{- 2} , h = 3 m , s = 6 m , k = 2500 \frac{N}{m} , x = 0.3$,

We get ${\mu}_{k} = 0.394$