Answer:
The speed of center of mass is 3.64 m/s.
Explanation:
Given that,
Total mass of the group of particles, m = 35 kg
Total kinetic energy of group of particles, [tex]K_T= 313 J[/tex]
The kinetic energy relative to the center of mass is, K = 81 J
We need to find the speed of center of mass. We know that the total kinetic energy is equal to the sum of rotational kinetic energy and the rotational kinetic energy.
[tex]K_T=E+K\\\\E=K_T-K[/tex]
Since, [tex]E=\dfrac{1}{2}mv_c^2[/tex], [tex]v_c[/tex] is the speed of center of mass
[tex]\dfrac{1}{2}mv_c^2=K_T-K\\\\v_c=\sqrt{\dfrac{2(K_T-K)}{m}} \\\\v_c=\sqrt{\dfrac{2(313-81)}{35}}\\\\v_c=3.64\ m/s[/tex]
So, the speed of center of mass is 3.64 m/s.
The speed of the center of mass will be "3.64 m/s".
According to the question,
Total mass, m = 35 kg
Total Kinetic energy, [tex]K_T[/tex] = 313 J
K.E relative to the mass, K = 81 J
We know the relation,
→ Total kinetic energy ([tex]K_T[/tex]) = E + K ...(Equation 1)
or,
E = [tex]K_T[/tex] + K ...(Equation 2)
here,
E = [tex]\frac{1}{2}[/tex]m[tex]v_c[/tex]²
By substituting "E" in "Equation 2",
→ [tex]\frac{1}{2}[/tex]m[tex]v_c[/tex]² = [tex]K_T[/tex] - K
[tex]v_c[/tex] = [tex]\sqrt{\frac{2(K_T-K)}{m} }[/tex]
By substituting the values,
= [tex]\sqrt{\frac{2(313-81)}{35} }[/tex]
= 3.64 m/s
Thus the above answer is correct.
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