Answer:
Explanation:
a. The compression over the other spring depends of the force that the first spring applies to the block
[tex]F=kx[/tex]
[tex]F=kx=(560\frac{N}{m})(0.1m)=56N[/tex]
This force allow us to calculate the compression in the other spring
[tex]x'=\frac{F}{k'}=\frac{56N}{377\frac{N}{m}}=0.14m[/tex]
b. We use the expression for the maximum velocity
[tex]v=A\omega[/tex]
where A is the amplitude 0.1m and w is the angular frecuency, which is calculated
[tex]\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{560\frac{N}{m}}{0.15kg}}=61.10 s^{-1}[/tex]
thus
[tex]v=(0.1m)(61.10s^{-1})=6.1\frac{m}{s}[/tex]
c. Here it is necessary to take into account the change in the kinetic energy due to the work done by the friction force
[tex]E_{k}=\frac{mv^{2}}{2}=2.79J\\f_{k}x=\mu N x= (0.5)(0.15kg*9.8\frac{m}{s^{2}})(1m)=0.735J\\[/tex]
The energy of the block at the moment of starting to compress the other spring is
[tex]E_{k}-f_{k}x=2.05J[/tex]
This energy will be potential energy of the other spring, hence
[tex]2.05J=U_{k'}=\frac{kx^{2}}{2}[/tex]
and by taking x' of this las expression we have
[tex]x'=\sqrt{\frac{2(2.05J)}{k'}}=\sqrt{\frac{2(2.05J)}{377\frac{N}{m}}}=0.10m[/tex]
I hope this is useful for you
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