Question

A spring compressed by 0.10 m from its rest length launches 0.53-kg block (starting at rest) along a rough horizontal surface with the coefficient of kinetic friction 0.24. If the spring constant is 515 N/m, how far will the block slide from its initial position?

Answer 1

Answer: $2.06\ m$

Explanation:

Given

Spring is compressed by 0.1 m from its rest length

Mass of block is 0.53 kg

The coefficient of kinetic friction is $\mu_k=0.24$

Spring constant $k=515\ N/m$

Here, the spring energy is consumed against work done by friction force

Friction force acting on the mass is $F=\mu_kmg$

Writing Elastic potential energy equals work done by friction force

$\Rightarrow \dfrac{1}{2}kx^2=\mu_kmg\cdot s\\\\\Rightarrow 0.5\times 515\times 0.1^2=0.24\times 0.53\times 9.8\cdot s\\\Rightarrow s=2.06\ m$

So, the block will slide up to a distance of $2.06\ m$