Question

1. -/2 points a block of mass 3.60 kg is placed against a horizontal spring of constant k = 735 n/m and pushed so the spring compresses by 0.0750 m. hint (a) what is the elastic potential energy of the block - spring system (in j)? j (b) if the block is now released and the surface is frictionless, calculate the block’s speed (in m/s) after leaving the spring. m/s resources read it

Explanation:

Step1: Calculate elastic potential energy

The formula for elastic potential energy is $U = \frac{1}{2}kx^{2}$, where $k$ is the spring - constant and $x$ is the compression. Given $k = 735\ N/m$ and $x=0.0750\ m$.
$U=\frac{1}{2}\times735\times(0.0750)^{2}$
$U=\frac{1}{2}\times735\times0.005625$
$U = 2.07\ J$

Step2: Use conservation of energy

The elastic potential energy is converted into kinetic energy when the block is released. $K=\frac{1}{2}mv^{2}=U$. We know $U = 2.07\ J$ and $m = 3.60\ kg$.
$\frac{1}{2}mv^{2}=2.07$
$v^{2}=\frac{2\times2.07}{3.60}$
$v^{2}=\frac{4.14}{3.60}=1.15$
$v=\sqrt{1.15}\approx1.07\ m/s$

Answer:

(a) $2.07$
(b) $1.07$