# An object of mass m is moving with a constant velocity v along a straight path. Find the work that needs to be done to stop it.

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The work-energy theorem states that the work done by external force on an object equals the change of the kinetic energy of the object:

`W = Delta K = K_f - K_i` . Here, `K_i` is the initial kinetic energy and `K_f ` is the final kinetic energy.

The kinetic energy is the energy associated with motion of an object. It is proportional to the mass and the square of the speed of the object:

`K = (mv^2)/2`

In the situation described in the problem, the object is initially moving with velocity v. Since its mass is m, the initial kinetic energy of the object is `K_i = (mv^2)/2` .

If the object stops after some work W is done on it, its velocity becomes 0, so its final kinetic energy becomes `K_f = 0` . Then, according to the work-energy theorem,

`W = K_f - K_i = 0 - (mv^2)/2=-(mv^2)/2` .

So the work that needs to be done to stop the object is equal to `-(mv^2)/2` . Notice that this is a negative value, because the kinetic energy decreases as a result of this work.