A load weighing 1 kg begins to be lifted by the rope vertically upwards c. constant acceleration.

A load weighing 1 kg begins to be lifted by the rope vertically upwards c. constant acceleration. For 2 with a rope tension force, the work was performed 48 J. Find the acceleration of the load.

m = 1 kg.

g = 9.8 m / s ^ 2.

t = 2 s.

A = 48 J.

а – ?

The work of force A is determined by the formula: A = F * S * cosα, where F is the pulling force of the rope, S is the movement of the load, ∠α is the angle between force F and displacement S.

∠α = 0 °.

S = h.

Let’s write Newton’s 2 law for the motion of bodies: m * a = F – m * g, where a is the acceleration of the body, F is the force that lifts the load, m * g is the force of gravity.

F = m * a + m * g = m * (a + g).

A = m * (a + g) * h * cosα.

h = a * t ^ 2/2.

A = m * (a + g) * a * t ^ 2 * cosα / 2 = m * a ^ 2 * t ^ 2 * cosαa / 2 + g * a * t ^ 2 * cosα / 2.

48 = 2 * a ^ 2 + 20 * a.

a ^ 2 + 10 * a – 24 = 0.

D = √ (100 – 4 * (- 24)) = √196 = 14.

a1,2 = (-10 ± 14) / 2.

a1 = – 12 has no physical meaning.

a2 = 2 m / s ^ 2.

Answer: the acceleration of the body is a = 2 m / s ^ 2.