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Sagot :
Given:
mass of the spike is
[tex]0.45\text{ kg}[/tex]initial speed od the spike is
[tex]v_=\text{ 1.9 m/s}[/tex]if the tie and spike together absorb 22 percent of the spike's initial kinetic energy.
Required:
calculate the increase in internal energy of the tie and spike.
Explanation:
here we apply conservation of energy.
[tex]\Delta U+\Delta K+\Delta P=0[/tex]total change in energy is zero.
here potential energy is not given so neglect that part. we have only
[tex]\Delta U+\Delta K=0[/tex][tex]\Delta U=-\Delta K[/tex]Here,
[tex]\Delta U=-(K_2-K_1)[/tex]K1 is initial kinetic energy and K2 is final kinetic energy that is equal to zero.
now we have
[tex]\Delta U=K_1[/tex]we know that kinetic energy is
[tex]K=\frac{1}{2}mv^2[/tex]then
[tex]\Delta U=\frac{1}{2}mv^2[/tex]plugging all the values in the above relation. we get
[tex]\begin{gathered} \Delta U=\frac{1}{2}\times0.45\text{ kg }\times(1.9\text{ m/s})^2\times0.22 \\ \Delta U=0.18\text{ J} \end{gathered}[/tex]Thus, the change in internal energy is
[tex]0.18\text{ J}[/tex]
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