Question

A 2.0 kg block is held at rest against a spring with a force constant k = 395 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 2.5 m/s.

Answer 1

Answer:

19.237 cm

Explanation:

m = Mass of block = 2 kg

v = Velocity of block = 2.5 m/s

$\mu$ = Coefficient of friction = 0.54

g = Acceleration due to gravity = 9.81 m/s²

d = Distance

k = Spring constant = 395 N/m

The kinetic energy of the block is after covering the patch is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}2\times 2.5^2\\\Rightarrow K=6.25\ J$

Work done by fricition force is given by

$W=f\Delta xcos \theta\\\Rightarrow W=\mu mg\Delta xcos \theta\\\Rightarrow W=0.54\times 2\times 9.81\times 0.1\times cos 180\\\Rightarrow W=-1.05948\ J$

Applying conseravation of energy

$K_i+U_i+W=K_f+U_f\\\Rightarrow 0+\dfrac{1}{2}kd^2-1.05948=6.25+0\\\Rightarrow d=\sqrt{\dfrac{(6.25+1.05948)2}{395}}\\\Rightarrow d=0.19237\ m$

The distance in centimeters if the spring was compressed was 19.237 cm