Question

What is the answer for a, b, c and d?

Answer 1

Given

m: mass

m = 1.20 kg

μ: kinetic friction

μ = 0.25

vi: speed contact light spring

vi = 3.40 m/s

k: spring costant force

k = 50 N/m

Procedure

a) distance of compression

energy balance equation

$\begin{gathered} \mu mg\cdot d+\frac{1}{2}kd^2-\frac{1}{2}mv^2_i=0 \\ 0.25\cdot1.20\cdot9.8\cdot d+\frac{1}{2}\cdot50\cdot d^2-\frac{1}{2}\cdot1.2\cdot3.40^2=0 \\ 2.94d+25.0d^2-6.936=0 \\ 25d^2+2.94d-6.936=0 \end{gathered}$

using the quadratic formula we get,

$d=0.471m$

(b) find the speed v, at the unstretched

energy balance equation

$\begin{gathered} \mu mg\cdot d+\frac{1}{2}mv^2_i=\frac{1}{2}kd^2 \\ \frac{1}{2}kd^2-\mu mg\cdot d=\frac{1}{2}mv^2 \\ \frac{1}{2}\cdot50\cdot0.471-0.25\cdot1.20\cdot9.8\cdot0.471=\frac{1}{2}\cdot1.20\cdot v^2 \\ 4.1612=0.6v^2 \\ v=\sqrt[]{\frac{4.1612}{0.6}} \\ v=2.63m/s \end{gathered}$

(c) Find the distance D where come to rest

energy balance equation

$\begin{gathered} \frac{1}{2}mv^2=\mu mg\cdot d \\ d=\frac{1}{2}\cdot\frac{mv^2}{\mu mg} \\ d=\frac{1}{2}\cdot\frac{v^2}{\mu g} \\ d=\frac{1}{2}\cdot\frac{2.63^2}{0.25\cdot9.8} \\ d=1.41m \end{gathered}$

(d) What if? The object becomes attached to the end of the spring

$\begin{gathered} \frac{1}{2}mv^2=\mu mg\cdot d+\frac{1}{2}kd^2 \\ \frac{1}{2}kd^2+\mu gm\cdot d-\frac{1}{2}mv^2=0 \\ 25d^2+2.94d-4.15=0 \\ d=\text{0}.35m \end{gathered}$

The distance would be 0.35m