Work  (Page 2/3)

It is, therefore, a good idea that we specify the force(s) that we have considered in the calculation of work. To appreciate this point, we consider a block being slowly raised vertically by hand to a height "h" as shown in the figure. As the block is not accelerated, the normal force applied by the hand is equal to the weight of the body :

$$\begin{array}{l}N=mg\end{array}$$

For gravity (gravitational force due to Earth), the force and displacement are opposite. Hence, work by gravity is negative.

The work done by gravity is :

$$\begin{array}{l}\Rightarrow W_G=-\left(mg\right)h=-mgh\end{array}$$

For normal force applied by the hand on the block, force and displacement are both in the same direction. Hence, work done is positive :

The work by the hand is :

$$\begin{array}{l}\Rightarrow W_H=NXh=mgh\end{array}$$

Thus, two works are though equal in magnitude, but opposite in sign. It can be easily inferred from the example here that work is positive, if both displacement and component of force along displacement are in the same direction; otherwise negative. It is also pertinent to mention that a subscripted notation for work as above is a good practice to convey the context of work. Finally, we should also note that net force on the body is zero. Hence, work by net force is zero - though works by individual forces are not zero.

Examples

In the discussion above, we have made two points : (i) the sign of work can be evaluated either evaluating "cosθ" or by examining the relative directions of the component of force and displacement and (ii) work is designated to named force. Here, we select two examples to illustrate these points. First example shows computation of work by friction - one of the forces acting on the body. The determination of sign of work is based on evaluation of cosine of angle between force and displacement. Second example shows computation of work by gravity. The determination of sign is based on relative comparison of the directions of the component of force and displacement.

Evaluation of cosine of angle

Problem 1 : A block of 2 kg is brought up from the bottom to the top along a rough incline of length 10 m and height 5 m by applying an external force parallel to the surface. If the coefficient of kinetic friction between surfaces is 0.1, find work done by the friction during the motion. (consider, g = 10 $m/s^2$ ).

Solution : We see here that there are four forces on the block : (i) weight (ii) normal force (iii) friction and (iv) force, "F" parallel to incline. The magnitude of external force is not given. We are, however, required to find work by friction. Thus, we need to know the magnitude of friction and its direction. As the block moves up, kinetic friction acts downward. Here, displacement is equal to the length of incline, which is 10 m.

From the figure, it is clear that friction force is given as :

$$\begin{array}{l}{F}_k={\mu }_{k}N={\mu }_{k}mg\cos \theta \\ \text{Here,}\sin \theta =\frac{5}{10}\\ \text{and}\cos \theta =\sqrt{(1-{\sin }^2\theta )}\\ \Rightarrow F=0.1x2x10x\sqrt{(1-\frac{5^2}{{10}^2})}=2x0.866=1.732N\end{array}$$

To evaluate work in terms of "Frcosφ", we need to know the angle between force and displacement. In this case, this angle is 180° as shown in the figure below.