Answer to Question #186963 in Mechanics | Relativity for crystal r

Question #186963

assume the spring constant in this lab is 10 N/m. if a mass of 1 lb is attached and the mass pulls for 5 inches. when the mass is released what will be its max minim speed going up in m/s,

and what will be it’s maximum acceleration in m/s?


1
Expert's answer
2021-04-29T10:41:10-0400


If we choose downward as the positive Y direction, then Newton's second law gives


Fy=may\sum{F_y }=ma_y

where


Fy=ky+mg=k(y1+y0)+mg\sum{F_y }=-ky+mg=-k(y_1+y_0)+mgmay=md2ydt2={y0=Const}=md2y1dt2ma_y=m\frac{d^2y}{dt^2}=\big\{ y_0=Const\big\}=m\frac{d^2y_1}{dt^2}ky0=mgequilibrium positionky_0=mg- equilibrium \space position

The equation of motion of the mass takes the form:


md2y1dt2=ky1m\frac{d^2y_1}{dt^2}=-ky_1

It has the solution:


y1(t)=y1maxcos(kmt)y_1(t)=y_{1max} cos\bigg (\sqrt{\frac{k}{m}}t\bigg)


Then, the speed of movement of the mass:


v1(t)=dy1dt=y1maxkmsin(kmt)v_1(t)=\frac{dy_1}{dt}=-y_{1max}\sqrt{\frac{k}{m}}sin{\bigg(\sqrt{\frac{k}{m}}t}\bigg)

The acceleration:


a1(t)=d2y1dt2=y1max(km)cos(kmt)a_1(t)=\frac{d^2y_1}{dt^2}=-y_{1max}\bigg(\frac{k}{m}\bigg)cos{\bigg(\sqrt{\frac{k}{m}}t}\bigg)

in SI units:


m=1lb=0.4536kg, y1max=5in=0.0254mm=1lb=0.4536kg, \space y_{1max}=5in=0.0254m

So, the maximum acceleration and maximum speed of the mass upward, taking into account the direction of the Y-axis:


v1max=y1maxkm=(0.0254m)10Nm0.4536kg0.119msv_{1max}=-y_{1max}\sqrt{\frac{k}{m}}=-(0.0254m)\sqrt{\frac{10 \frac N m}{0.4536kg}}\approxeq-0.119 \frac m sa1max=y1max(km)=(0.0254m)(10Nm0.4536kg)0.56m2sa_{1max}=-y_{1max}\bigg({\frac{k}{m}}\bigg)=-(0.0254m)\bigg({\frac{10 \frac N m}{0.4536kg}}\bigg)\approxeq-0.56 \frac {m^2} s


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