If the work required to stretch a spring 3 ft beyond its natural length is 15 ft-lb, how much work is needed to stretch it 27 in. beyond its natural length?

8.4 ft-lb

Explanation:

Work = change in energy

W = ½ kx²

When x = 3 ft, W = 15 ft-lb:

15 ft-lb = ½ k (3 ft)²

k = 30/9 lb/ft

When x = 27 in = 2.25 ft:

W = ½ kx²

W = ½ (30/9 lb/ft) (2.25 ft)²

W = 8.4375 ft-lb

Rounding to 2 sig-figs, it takes 8.4 ft-lb of work.

see earlier

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