Work  (Page 3/3)

$$\begin{array}{l}W=Fr\cos \phi \end{array}$$

$$\begin{array}{l}\Rightarrow W_F=Fr\cos \phi =1.732x10x\cos {180}^0\\ \Rightarrow W_F=1.732x10x(-1)=-17.32J\end{array}$$

This example brings out the concept of work by named force (friction). The important point to note here is that we could calculate work by friction even though we were not knowing the magnitude of force of external force, "F". Yet another point to note here is that computation of work by friction is actually independent of - whether block is accelerated or not? In addition, this example illustrates how the evaluation of the cosine of angle between force and displacement determines the sign of work.

Relative comparison of directions

Problem 2 : A block of 2 kg is brought up from the bottom to the top along a smooth incline of length 10 m and height 5 m by applying an external force parallel to the surface. Find work done by the gravity during the motion. (consider, g = 10 $m/s^2$ ).

Solution : In this problem, incline is smooth. Hence, there is no friction at the contact surface. Now, the component of gravity (weight) along the direction of the displacement is :

$$\begin{array}{l}\Rightarrow mg\sin \theta =2x10x\frac{5}{10}=10N\end{array}$$

We, now, determine the magnitude of work without taking into consideration of sign of work. The magnitude of work by gravity is :

$$\begin{array}{l}\Rightarrow W_G=mg\sin \theta xr=10x10=100J\end{array}$$

Once magnitude is calculated, we compare the directions of the component of force and displacement. We note here that the component of weight is in the opposite direction to the displacement. Hence, work by gravity is negative. As such, we put a negative sign before magnitude.

$$\begin{array}{l}\Rightarrow W_g=-100J\end{array}$$

It is clear that this second method of computation is easier of two approaches. One of the simplifying aspect is that we need to calculate cosine of only acute angle to determine the magnitude of work without any concern about directions. We assign sign, subsequent to calculation of the magnitude of work.

Work in three dimensions

We can extend the concept of work to motion in three dimensions. Let us consider three dimensional vector expressions of force and displacement :

$$\begin{array}{l}\mathbf{F}=F_x\mathbf{i}+F_y\mathbf{j}+F_z\mathbf{k}\end{array}$$

and

$$\begin{array}{l}\mathbf{r}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\end{array}$$

The work as dot product of two vectors is :

$$\begin{array}{l}W=\mathbf{F}\mathbf{.}\mathbf{r}=(F_x\mathbf{i}+F_y\mathbf{j}+F_z\mathbf{k})\mathbf{.}(x\mathbf{i}+y\mathbf{j}+z\mathbf{k})\\ W=F_{x}x+F_{y}y+F_{z}z\end{array}$$

Form the point of view of computing work, we can calculate "work" as the sum of the products of scalar components of force and displacement in three mutually perpendicular directions along the axes with appropriate sign. Since respective components of force and displacement are along the same direction, we can determine work in each direction with appropriate sign. Finally, we compute their algebraic sum to determine work by force, "F".

Work by a variable force

We have defined work for constant force. This condition of constant force is, however, not a limitation as we can use calculus to compute work by a variable force. In order to keep the derivation simple, we shall consider force and displacement along same straight line or direction. We have already seen that calculation of work in three dimensional case is equivalent to calculation of work in three mutually perpendicular directions.

A variable force can be approximated to be a series of constant force of different magnitude as applied to the particle. Let us consider that force and displacement are in the same x-direction.

For a given small displacement (Δx), let $$F_x$$ be the constant force. Then, the small amount of work for covering a small displacement is :

$$\begin{array}{l}\Delta W=F_x\Delta x\end{array}$$

We note that this is the area of the small strip as shown in the figure above. The work by the variable force over a given displacement is equal to sum of all such small strips,

$$\begin{array}{l}W=\sum \Delta W=\sum F_x\Delta x\end{array}$$

For better approximation of the work by the variable force, the strip is made thinner as Δx-->0, whereas the number of stips tends to be infinity. For the limit,

$$\begin{array}{l}W={lim}_{\Delta x\to 0}\sum F_x\Delta x\end{array}$$

This limit is equal to the area of the plot defined by the integral of force function F(x) between two limits,

$$\begin{array}{l}W={\int }_{x_1}^{x_2}F\left(x\right)đx\end{array}$$

Example

Problem 3 : A particle moves from point A to B along x - axis of a coordinate system. The force on the particle during the motion varies with displacement in x-direction as shown in the figure. Find the work done by the force.

Solution : The work done by the force is :

$$\begin{array}{l}W={\int }_{x_1}^{x_2}F\left(x\right)đx=\text{Area between plot and x-axis within the limits}\end{array}$$

Now, the area is :

$$\begin{array}{l}W=-\frac{1}{2}x0.5x15+\frac{1}{2}x0.5x20+\frac{1}{2}x1x(30+20)+\frac{1}{2}x2x(30+10)\\ W=-3.75+5+25+40=66.25J\end{array}$$

Acknowledgment

Author wishes to thank Keith for making suggestion to remove calculation error in the module.