# The load is lifted up on the cord with an acceleration of 6 m / s2, then lowered with the same

**The load is lifted up on the cord with an acceleration of 6 m / s2, then lowered with the same acceleration, determine the ratio of the cord elongations**

a = 6 m / s2.

g = 9.8 m / s2.

x1 / x2 -?

Let’s write Newton’s 2 law for a load: m * a = N + m * g, where m is the mass of the load, a is the acceleration of the load, N is the elastic force of the cord, m * g is the force of gravity.

According to Hooke’s law: N = k * x, where k is the stiffness of the cord, x is the elongation of the cord.

Let’s write 2 Newton’s law for projections on the vertical axis OU directed upwards.

1) For movement with acceleration a directed upwards.

m * a = N – m * g.

m * a = k * x1 – m * g.

k * x1 = m * (a + g).

x1 = m * (a + g) / k.

2) For movement with acceleration a directed downward.

– m * a = N – m * g.

– m * a = k * x2 – m * g.

k * x2 = m * (g – a).

x2 = m * (g – a) / k.

x1 / x2 = m * (a + g) * k / k * m * (g – a) = (a + g) / (g – a).

x1 / x2 = (9.8 m / s2 + 6 m / s2) / ((9.8 m / s2 – 6 m / s2) = 4.2.

Answer: x1 / x2 = 4.2.