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

19.237 cm

Step-by-step 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=(1)/(2)mv^2\\\Rightarrow K=(1)/(2)2* 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* 2* 9.81* 0.1* cos 180\\\Rightarrow W=-1.05948\ J`

Applying conseravation of energy

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

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