When lifting a load weighing 30 kg, a work of 3.2 kJ was performed. The load was lifted from rest at a uniform acceleration

When lifting a load weighing 30 kg, a work of 3.2 kJ was performed. The load was lifted from rest at a uniform acceleration to a height of 10 m. With what acceleration was the load lifted?

m = 30 kg.

A = 3.2 kJ = 3200 J.

g = 9.8 m / s2.

h = 10 m.

a -?

The work of the force A, which will lift the load, is expressed by the formula: A = F * S * cosα, where F is the force that lifts the load, S is the movement of the load, ∠α is the angle between F and S.

∠α = 0 °, cos0 ° = 1.

S = h.

A = F * h.

F = A / h.

Let us write Newton’s 2 law when lifting a body: m * a = F – m * g = A / h – m * g.

The formula for determining the acceleration will be: a = A / h * m – g.

a = 3200 J / 10 m * 30 kg – 9.8 m / s2 = 0.86 m / s2.

Answer: the body was lifted with an acceleration of a = 0.86 m / s2.