A body weighing 4 kg initially resting on a smooth surface is acted upon by a force F of a horizontal direction

A body weighing 4 kg initially resting on a smooth surface is acted upon by a force F of a horizontal direction equal to 2 N in absolute value for 3 s. The work of force for the specified time is equal to?

Given:

m = 4 kilograms – the mass of a certain body;

F = 2 Newton – the force that acts on the body;

t = 3 seconds – the period of time during which the action of the force lasted.

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

According to the condition of the problem, the friction force is not taken into account in the calculation.

Let’s find the acceleration with which the body was moving:

a = F / m = 2/4 = 0.5 m / s2.

Let’s find the path traversed by the body:

s = a * t ^ 2/2 = 0.5 * 3 ^ 2/2 = 0.5 * 9/2 = 2.25 meters.

Then, the work will be equal to:

A = F * s = 2 * 2.25 = 4.5 Joules.

Answer: in 3 seconds, a work equal to 4.5 Joules was done by force.