Posted by Anonymous on Sunday, April 3, 2011 at 10:06pm.
(a) MgX - (1/2)kX^2 = increase in kinetic energy = (1/2) M V^2 + (1/2) I w^2
Solve for V
M = 5 kg
g = 9.8 m/s^2
X = 1 m
k = 35 N/m
I = (1/2)*Mpulley)*Rpulley^2
w = V/Rpulley
b) Maximum extension Xmax occurs when V = 0
Related Questions
physics/ torques - A block of mass m = 5 kg is attached to a spring (k = 35 N/m...
physics - A block of mass m = 5 kg is attached to a spring (k = 28 N/m) by a ...
physics - A block of mass m = 2.67 kg is attached to a spring (k = 32.3 N/m) by ...
Physics - A block of mass m = 3 kg is attached to a spring (k = 28 N/m) by a ...
i need help i couldnt figure this out... - A block of mass m = 3 kg is attached ...
physics - A block of mass m = 3 kg is attached to a spring (k = 28 N/m) by a ...
physics - A block with mass m =7.3 kg is hung from a vertical spring. When the ...
PHYSICS - A block is on a 24 degree incline, with a rope attached to a spring at...
physics - A light rope is attached to a block with a mass of 6 kg that rests on ...
Physics - A 1.7kg block and a 2.7kg block are attached to opposite ends of a ...
For Further Reading