Question

A body is at rest acted upon by a force f. It acquires speed v and possesses acceleration a. Find the work done.

Answer 1

As the work done by the body is asked, let us figure out the total force acting on the body and total energy of the system. As work done is equal to change in kinetic energy acting on the body, we can find the work done by finding the total energy easily.

Formula used:
$\begin{aligned} & W=T.E \\\ & W=F.s \\\ \end{aligned}$

Complete step by step answer:
Let us find the kinetic energy acting on the body. As it is given in the question that the body acquired a velocity v, the total energy of the body is equal to the kinetic energy of the body,
Kinetic energy of the body is equal to,
$K.E=\dfrac{1}{2}m{{v}^{2}}...(1)$
Also, the force acting on the body is equal to,
$\begin{aligned} & F=ma \\\ & \Rightarrow m=\dfrac{F}{a}...(2) \\\ \end{aligned}$
As the mass of the body is not given in question, substitute equation 2 in equation 1, we greet,
$\begin{aligned} & K.E=\dfrac{1}{2}\times \dfrac{F}{a}\times {{v}^{2}} \\\ & \Rightarrow K.E=\dfrac{F{{v}^{2}}}{2a} \\\ \end{aligned}$
As, the work done by a body is equal to the change in kinetic energy of the body according to the work-energy theorem,
$\begin{aligned} & W=\Delta K.E \\\ & W=\dfrac{F{{v}^{2}}}{2a} \\\ \end{aligned}$
Hence, the above is the formula for work done in known terms.

Additional Information: The principle of work and kinetic energy also known as work energy theorem states that the work done by the sum of all forces acting on a particle is equal to the change in the kinetic energy of the particle. This can also be extended to rigid bodies by defining the work of the torque and rotational kinetic energy. According to this theorem, when an object slows down, its final kinetic energy is less than its initial kinetic energy, the change in its kinetic energy is negative, and so is the net work done on it. If an object speeds up, the net work done on it is positive. When calculating the network, you must include all the forces that act on an object. If you leave out any forces that act on an object, or if you include any forces that do not act on it, you will get a wrong result.

Note: The work done on any object is not always positive. Negative work removes or dissipates energy from the system. For example, while pulling a book they are longer on the surface at constant velocity the work done is positive. The friction force that opposes the displacement of the book he's doing negative work here. Hey there for the work done can be negative too.