Question

A 1.00 kg object slides to the right on a surface with kinetic friction coefficient of 0.25. The initial speed of the object when it makes contact with a light spring
(kspring =50.00 N/m) is 3.00 m/s. The object comes to rest after the spring has been compressed a distance d. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring’s un-stretched position. Finally, the object comes to rest a distance D to the left of the un-stretched spring.
Find d, D , and the speed v at the un-stretched position when the object is moving to the left.

Answer 1

Answer on Question # 83382, Physics / Mechanics | Relativity

Question 1. A $m=1\,kg$ object slides to the right on a surface with kinetic friction coefficient of $\mu=0.25$. The initial speed of the object when it makes contact with a light spring ($k$ spring $=50\,N/m$) is $v_{0}=3\,m/s$. The object comes to rest after the spring has been compressed a distance $d$. The object is then forced toward the left by the spring and continues to move in that direction beyond the spring’s un-stretched position. Finally, the object comes to rest a distance $D$ to the left of the un-stretched spring. Find $d,D$, and the speed $v$ at the un-stretched position when the object is moving to the left.

Solution. The kinetic energy of the object was spent on friction and compressing the spring. Friction did the work, $\mu mgd$ and the spring (elastic force) did the work $kd^{2}/2$. Hence,

$mv_{0}^{2}/2=\mu mgd+kd^{2}/2\Rightarrow d^{2}+2\mu mgd/k-mv_{0}^{2}=0\Rightarrow d=\frac{1}{2}\left(-\mu mg/k\pm\sqrt{(\mu mg/k)^{2}+mv_{0}^{2}/k}\right),$

$d>0\Rightarrow d=-0.25\cdot 9.8/50\pm\sqrt{(0.25\cdot 9.8/50)^{2}+3^{2}/50}\approx 0.378\,m.$

In the moment, when the spring at the un-stretched position, the spring did work which was spent on friction and the kinetic energy of the object. Then,

$kd^{2}/2=\mu mgd+mv^{2}/2\Rightarrow v^{2}=\left(kd^{2}+2\mu mgd\right)/m,$

$v>0\Rightarrow v=\sqrt{\left(kd^{2}-2\mu mgd\right)/m}=\sqrt{\left(50\cdot 0.378^{2}-2\cdot 0.25\cdot 9.8\cdot 0.378\right)}\approx 2.3\,ms^{-1}.$

When the object stopped, the kinetic energy was spent on friction. So,

$mv^{2}/2=\mu mgD\Rightarrow D=mv^{2}/(2\mu mg)=2.3^{2}/(2\cdot 0.25\cdot 9.8)\approx 1.08\,m.$

$\Box$