The speed of a body weighing 2 kg changes according to the equation υх = 5 + 4t + 2t ^ 2.

The speed of a body weighing 2 kg changes according to the equation υх = 5 + 4t + 2t ^ 2. Determine the work of the forces acting on the body during the first four seconds.

Given:

m = 2 kilograms is the mass of a moving body;

v (t) = 5 + 4 * t + 2 * t ^ 2 – dependence of speed on time;

t = 4 seconds – time span.

It is required to determine A (Joule) – the work of the resultant forces for the time interval t.

Let us find the initial velocity of the body at the moment of time t = 0:

v (0) = 5 + 4 * 0 + 2 * 0 ^ 2 = 5 m / s.

Let’s find the speed of the body at the moment of time t = 4 seconds:

v1 = 5 + 4 * 4 + 2 * 4 ^ 2 = 5 + 16 + 32 = 53 m / s.

Then the average acceleration will be:

a = (v1 – v0) / t = 53 – 5/4 = 48/4 = 12 m / s ^ 2.

The force acting on the body is equal to:

F = m * a = 2 * 12 = 24 Newtons.

During time t, the body has traveled a path equal to:

S = v0 * t + a * t ^ 2/2 = 5 * 4 + 12 * 4 ^ 2/2 = 20 + 96 = 116 meters.

Then the work will be equal to:

A = F * S = 24 * 116 = 2784 Joules.

Answer: the work will be equal to 2784 Joules.