When a 3.30-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.80 cm. If the 3.30-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it?

0.0127 m or 1.27 cm

Explanation:

From Hook's law,

F = ke Equation 1

Where F = Force on the spring, k = spring constant of the spring, e = extension of the spring.

Note: The Force of the spring is the weight of the mass hung on the spring,

I.e,

F = mg Equation 2

Where m = mass of the object hung on the spring, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke Equation 3

make k the subject of the equation

k = mg/e Equation 4

Given: m = 3.3 kg,  e = 2.8 cm = 0.028 m, g = 9.8 m/s²

Substitute into equation 4

k = 3.3(9.8)/0.028

k = 1155 N/m

When the 3.3 kg block is removed,

make e the subject of the equation

e = mg/k Equation 5

Given: m = 1.5 kg, g = 9.8 m/s², k = 1155 N/m

Substitute into equation 5

e = 1.5(9.8)/1155

e = 0.0127 m

e = 1.27 cm

-4 on

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