Answer:
a).1.147cm
b).1.0256 J
Explanation:
Using law of force and applications Law Hooke
[tex]F=k*d[/tex]
[tex]d=\frac{F}{k}[/tex]
But first have to know force and constant in initial conditions
[tex]F=3.4kg*9.8\frac{m}{s^{2}} \\F=33.32N\\k=\frac{F}{d}\\ d=2.6cm\frac{1m}{100cm}=0.026m=26x10^{-3}m[/tex]
[tex]k=\frac{33.23N}{26x10^{-3}}=1281.53 \frac{N}{m}[/tex]
a).
[tex]d=\frac{F}{k}[/tex]
[tex]d=\frac{1.5kg*9.8\frac{m}{s^{2}}}{1281.53 \frac{N}{m}}=\frac{14.7N}{1281.53 \frac{N}{m}}\\ d=0.0147m=11.47x10^{-3}m[/tex]
[tex]d=11.47x10^{-3}m*\frac{100cm}{1m} \\d=1.147cm[/tex]
b).
Wokr is the force done so using the equation
[tex]d=4cm*\frac{1m}{100cm}=0.04m[/tex]
[tex]Ew=\frac{1}{2}*k*d^{2}\\Ew=\frac{1}{2}*1281.53N*(0.04m)^{2} \\Ew=1.025N*m^{2} \\Ew=1.025 J[/tex]
