Question

A 1.5 kg box moves back and forth on a horizontal frictionless surface between two different springs as shown. The box is initially pressed against the stronger spring compressing it 4.0 cm, and then is released from rest. (a) By how much will the box compress the weaker spring? (b) What is the maximum speed the box will reach?

Answer 1

let k1,k2 are the spring constants of the two springs

then potential energy of the spring with comprression or elangation is 1/2 kx^2

here two spring system attached to a box of mass 1.5 kg, given compression on strong spring is 0.04 m

now 1/2 kx1^2= 1/2 kx2^2 ===> x2 = sqrt(k1/k2)* x1

                                                                 = sqrt(k1/k2)*0.04

now when it is released from rest total potential energy is converted in to kinetic energy

    1/2 (k1+k2)x^2= 1/2 m V^2

V= sqrt{(k+k2)/m}* x

the maximum speed the box will reach is V= sqrt{(k+k2)/m}* x