Question

A 2.0 kg block is held at rest against a spring with a force constant k = 264 N/m. The spring is compressed a certain distance and then the block is released. When the block is released, it slides across a surface that has no friction except for a 10.0 cm section that has a coefficient of friction μk = 0.54.

Find the distance in centimeters the spring was compressed such that the block's speed after crossing the rough area is 2.7 m/s.

Answer 1

Answer:

21.73 cm

Explanation:

We have given parameters:

Mass of block, m = 2.0 kg

Force constant of spring, k = 264 N/m

Length of rough area,  L  = 10 cm = 0.1 m

Co-efficient of kinetic friction , $\mu_{k}$ = 0.54

Block's speed after crossing rough area, v = 2.7 m/s

Block's initial speed ( when it was released from compressed spring), $v_{0}$ = 0 m/s

We need to find the distance that the spring was initially compressed, x = ?

Hence, we well apply Work-Energy principle which indicates that,

Work done by the friction = Change in the total energy of block

$- \mu_{k} * mg * L = (\frac{1}{2} * mv^{2} - \frac{1}{2} * mv_{0} ^{2} ) + (0 - \frac{1}{2} * kx^{2})$

-0.54 * 2 * 9.8 * 0.1 = (1/2 * 2 * $2.7^{2}$ - 1/2 * 2 * 0) + (0 - 1/2 * 264 * $x^{2}$)

x = 0.2173 m = 21.73 cm