A meter stick is found to balance at the 49.7-cm mark when placed on a fulcrum. When a 40.0-gram mass is attached at the 14.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick?

103 g

Explanation:

given that

Mass of the attached mass, m = 40 g

Initial distance of balance, x1 = 49.7 cm = 0.497 m

Final distance of balance, x2 = 39.2 cm = 0.392 m

Equilibrium point of attachment, x3 = 14 cm = 0.14 m

To get the mass of the meter stick, we use the relation

M = m.[(x2 - x3) / (x1 - x2)]

And on solving in full, we gave

M = 40.[(0.392 - 0.14) / [0.497 - 0.392)]

M = 40(0.258 / 0.105)

M = 40 * 2.457

M = 102.8 g

Therefore the needed work good for mass of the stick j

Is 103 g

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft