# A load weighing 60 kg is lifted upward with an acceleration of 2 m / s to find the work

**A load weighing 60 kg is lifted upward with an acceleration of 2 m / s to find the work of the body lifting to a height of 50 m.**

m = 60 kg.

g = 9.8 m / s ^ 2.

a = 2 m / s ^ 2.

h = 50 m.

A -?

The work of force A is determined by the formula: A = F * S * α, where F is the force that acts on the body, S is the movement of the body, ∠α is the angle between the force F and S.

A = F * h.

Let’s write 2 Newton’s law in vector form: m * a = F + m * g.

Let’s write 2 Newton’s law for projections on the vertical axis directed upwards: m * a = F – m * g.

The force F with which the load is lifted can be expressed by the formula: F = m * a + m * g = m * (a + g).

A = m * (a + g) * h.

A = 60 kg * (2 m / s ^ 2 + 9.8 m / s ^ 2) * 50 m = 35400 J.

Answer: the force F, which lifts the load, does the work A = 35400 J.