a) depends on the difference in the two masses.
energy: PE converted to KE:
(M-m)g*4.8
that PE is converted to ke of the two bodies. KE=1/2 M V^2+1/2 m V^2 set that equal to the PE, solve for V
b) knowing the 1/2 m V^2 at release, then the max height above release time can be found by max PE=mg*4.8+1/2 m V^2
then from max PE, =mgh, solve for h.
The masses are each initially 1.8m above the ground, and the massless frictionless pulley is 4.8m above the ground. (a) calculate the velocity of the lighter mass at the instance the heavier mass hits the ground. (b) find the maximum height reached by the lighter object (from the ground level) after the system is released. Use the energy method to find the answers.
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