Question

A $10kg$box slides $4.0m$ down the frictionless ramp shown inFIGURE CP10.72, then collides with a spring whose spring constant is $250N/m$.

a. What is the maximum compression of the spring?

b. At what compression of the spring does the box have its maximum speed?

Short answer

(a) The maximum compression of the spring $=1.46m$.

(b) When compressed$19.6cm$, the box has a maximum speed.

Step-by-step solution

Given information (part a)

$$\begin{gathered} \text{Weightofthebox}=10kg \\ \text{Thedistanceoftheslide}=4.0m \\ \text{Springconstant},k=250N/m \end{gathered}$$

Explanation (part a)

By using the conservation of energy equation.

$K_i+U_{si}+U_{gi}=K_f+U_{sf}+U_{gf}$

Kinetic energy at the initial and final positions is zero.

Therefore,

localid="1649400284183" $$\begin{gathered} U_{si}+U_{gi}=U_{sf}+U_{gf} \\ \frac{1}{2}k(s_i{)}^2+mg{y}_i=\frac{1}{2}m{v}_1^2+\frac{1}{2}k(s_f{)}^2+mg{y}_f \\ {y}_i=s_i\sin \left(30°\right) \\ {y}_f=s_f\sin \left(30°\right) \\ \frac{1}{2}k(0{)}^2+mg{s}_i\sin \left(30°\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ mg{s}_i\sin \left(30°\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ \left(10kg\right)\left(9.8m/s^2\right)\left(4m\right)\left(\frac{1}{2}\right)=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k(s_f{)}^2 \\ 196J=mg{s}_f\sin \left(30°\right)+\frac{1}{2}k\left(s_f{)}^2 \\ \frac{1}{2}\left(250N/m\right)\right(s_f{)}^2+\left(10kg\right)\left(9.8m/s^2\right)\left(\frac{1}{2}\right)s_f=196J \\ \left(125N/m\right)(s_f{)}^2+\left(49kgm/s^2\right)s_f=196J\cdots \cdots \left(1\right) \\ \text{Bysolvingtheequation(1)weget} \\ {\text{s}}_f\text{=1.46}m\text{and}-1.07m \end{gathered}$$

$\mathrm{The}\mathrm{maximum}\mathrm{compression}\mathrm{of}\mathrm{the}\mathrm{spring}\text{is}1.46\mathrm{m}$

Given information (part b)

Weight of the box$=10kg$

The distance of the slide$=4.0m$.

Spring constant, $k=250N/m$

Explanation (part b)

The net force at the initial contact s,

$F_s=-mg\sin \left(30°\right)$

The net force at the maximum displacement

$F_s=-ks-mg\sin \left(30°\right)$

The maximum speed at the compression of the spring at $F=0$is

localid="1649399872122" $$\begin{gathered} 0=-ks-mg\sin \left(30°\right) \\ s=\frac{-mg\sin \left(30°\right)}{k} \\ s=\frac{-\left(10kg\right)\left(9.8m/s^2\right)\left({\displaystyle \frac{1}{2}}\right)}{250N/m} \\ s=\frac{-49kg.m/s^2}{250N/m} \\ s=-0.196m \end{gathered}$$

At $19.6cm$the box has its maximum speed.