4.7 Work-Energy Theorem

4.7 Work-Energy Theorem

Whenever work is done on a body, it increases its energy. For example, if a force $F$ acts on a body of mass $m$, initially moving with velocity $v_{1}$, through a distance $d$ and increases its velocity to $v_{2}$, then the acceleration produced will be:

or

From the second law of motion:

or

Comparing Eqs. 4.16 and 4.17, we have:

or

This expression is the work-energy theorem. It states that:

The change in kinetic energy of an object is equal to the work done on it by a net force.

This is also known as work-energy principle.

The work-energy theorem is applicable for any direction of the force relative to the displacement. For instance, an object with kinetic energy can perform work if it is allowed to push or pull on another object. In this case, the work will be taken as negative and the kinetic energy of the object will decrease. The theorem remains valid even if the force may vary from point to point.

Example 4.4

A motorcycle rider weighing $60\,\mathrm{kg}$ is coasting down a $24^{\circ}$ slope. The weight of motorcycle is $30\,\mathrm{kg}$. At the top of the slope, the speed of motorcycle is $3.2\,\mathrm{m\,s^{-1}}$. If the kinetic frictional force is $100\,\mathrm{N}$, what will be the speed of the motorcycle $72\,\mathrm{m}$ downhill?

Solution

The normal force $F_{n}$ is balanced by the component of weight ($mg\cos 24^{\circ}$) perpendicular to the slope. Let the kinetic frictional force is $f$, then the net force $F$:

where $m =$ total mass $= 60\,\mathrm{kg} + 30\,\mathrm{kg} = 90\,\mathrm{kg}$

or

Work done: $W = Fd = 252.8\,\mathrm{N} \times 72\,\mathrm{m} = 18201.6\,\mathrm{J}$

As work is positive, so applying work-energy theorem:

From here: $(\mathrm{K.E})_{f} = W + (\mathrm{K.E})_{i}$

Putting the values, we have:

This gives: $v_{f}^{2} = 414.7\,\mathrm{m^{2}\,s^{-2}}$

or $\mathbf{v_{f} = \sqrt{414.7\,\mathrm{m^{2}\,s^{-2}}} = 20.4\,\mathrm{m\,s^{-1}}}$