Question #118574

A block of mass 2.01kg is at rest on a horizontal,frictionless surface. On one side a relaxed spring with a spring constant 192N/m connects it to a wall, on the other side of light string passing over a frictionless pulley connects it to the hanging block with a mass of 4.02kg. The pulley has a moment of inertia 0.006kg per square meter and a radius of 0.055m. A short time after the hanging block is released from rest it has fallen through a height of 0.09m.

a) determine the combined total kinetic energy of the blocks-and -pulley system,and

b)the rotational kinetic energy of the pulley.

a) determine the combined total kinetic energy of the blocks-and -pulley system,and

b)the rotational kinetic energy of the pulley.

Expert's answer

Corrected!

(a)

"m_1gx=KE+\\frac{kx^2}{2}\\to KE=m_1gx-\\frac{kx^2}{2}=4.02\\cdot9.81\\cdot0.09-\\frac{192\\cdot0.09^2}{2}=2.77J"

(b)

"KE=\\frac{m_1v^2}{2}+\\frac{m_2v^2}{2}+\\frac{I\\omega^2}{2}=\\frac{m_1+m_2}{2}\\omega^2r^2+\\frac{I\\omega^2}{2}"

"\\omega=\\sqrt{\\frac{KE}{\\frac{m_1+m_2}{2}r^2+\\frac{I}{2}}}=\\sqrt{\\frac{2.77}{\\frac{4.02+2.01}{2}0.055^2+\\frac{0.006}{2}}}=15.12rad\/s"

"KE_p=\\frac{I\\omega^2}{2}=\\frac{0.006\\cdot15.12^2}{2}=0.68J"

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