Question

1. A 220 g stone is dropped onto a relaxed vertical spring that has a spring constant of k = 4.5 N/cm. The stone becomes attached to the spring and compresses the spring 17 cm before momentarily stopping. While the spring is being compressed, what work is done on the stone by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the stone just before it hits the spring? (Assume no friction) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

Answer 1

x = 0.17 m

a) work done by gravitational force is

W_g = m*g*x

using known values,

W_g = 0.22 * 9.8 * 0.17

W_g = 0.367 Joules

b) work done by spring,

W_s = (1/2)*k*x²

where k is spring constant

W_s = 0.5 * 450 * 0.17²

W_s = - 6.5 Joules

work done by spring to stop stone.

c) using work energy principle,

net work done = change in kinetic energy

W_g + W_s = Kf - Ko

as the stone stops moving after 0.17 m then final kinetic energy is 0.

0.367 - 6.5 = 0 - (0.5 * m * Vo²)

solving for Vo,

Vo = 7.47 m/s

d) if initial velocity is doubled, Vo = 14.94 m/s

using work energy principle again,

m*g*h + (0.5*k*h²) = - 0.5 * m * Vo²

using all values,

0.22 * 9.8 * h + (0.5* 450 * h²) = 0.5 * 0.22* 14.94²

225*h² + 2.16*h - 24.55 = 0

solving for h,

h = 0.325 m

the spring will compress 32.5 cm