Question

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.

Answer 1

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)

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

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

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