Question

SQU is trying a new particle accelerator design with a mechanical component using a spring that does not obey Hooke's law. When stretched/compressed x metres, it exerts a force of 38.6x3+40.6x (a) Compute the work required to stretch the spring from x=0.8 m to x=2.3 m (b) With one end of the spring fixed, a particle of mass 2.2 kg is attached to the other end of the spring when it is stretched by an amount x -2.3 m If the particle is then released from rest, what is its speed at the instant the stretch in the spring is x -0.8 m? m/s (c) is the force exerted by the spring conservative or nonconservative Explain Previous page Next page CATION

Answer 1

the spring force is .

spring = 38.6.0 + 40.6.2

a) Wok done by the spring force during strech from x=0.8 m to x=2.3 m, is [answer].

2.3 2.3 -Fspringda (38.6x2 + 40.6x) dx = -360.49J 10.8 J0.8

The work done is negative, because the force acts towards mean position and displacement is away from mean position.

b) given : mass = m = 2.2 kg

When the particle is released at x = 2.3 m, the work done by the spring force will be positive and this work done will provide kinetic energy to the body connected to it.

The work done by the spring force from x=2.3 m to x = 0.8 m, is 360.49 J.

Therefore, from work energy principle :

work done = change in kinetic energy

=> [at x = 2.3 m, the body is released, i.e; it's velocity is zero at x = 2.3 m]

360.49 = -mu

=>

360.49 = * 2.2 * U

=> . [answer]

U = 360.49 1.1 = 18.10m/s

c) The force exerted by spring is conservative , because, while stretching or compressing, the work done is negative. and while relaxing or while mi=oving towards mean position it's work done is positve.