# What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0190 m). See details.

## A 2.55 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has a force constant of 815 N/m. The coefficient of kinetic friction between the floor and the block is 0.35. The block and spring are released from rest and the block slides along the floor.

Apr 11, 2017

$v = 0.32 \text{ } \frac{m}{s}$

#### Explanation: $\text{When a spring is compressed by a Force of F,The potential energy }$$\text{is stored on it.}$

$\text{if the spring is released, the Potential energy turns into kinetic energy.}$

$\text{The object is repulsed by the call-back force which occurs on the spring}$

$F ' = K \cdot \Delta x$

$K = 815 \frac{N}{m}$
$\Delta x = 0.0190 \text{ } m$

$F ' = 815 \cdot 0.019 = 15.48 \text{ } N$

${F}_{f} = k \cdot m \cdot g$

${F}_{f} = 0.35 \cdot 2.55 \cdot 9.81 = 8.76 N \text{ Friction force}$

$\text{acceleration of object is :}$

$a = \frac{F ' - {F}_{f}}{m}$

$m = 2.55 k g \text{ mas of object}$

$a = \frac{15.48 - 8.76}{2.55}$

$a = 2.64 \text{ } \frac{m}{s} ^ 2$

${v}^{2} = 2 \cdot a \cdot \mathrm{di} s p l a c e m e n t$

${v}^{2} = 2 \cdot 2.64 \cdot 0.0200$

${v}^{2} = 0.1056$

$v = \sqrt{0.1056}$

$v = 0.32 \text{ } \frac{m}{s}$