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These results are exactly the same as for the vertical motion discussed in the previous section. Interestingly, work along any incline that reaches same height is same. The objects released from a given height will have same velocity and kinetic energy, irrespective of the length of incline involved.
The work, kinetic energy and velocity are independent of path between end points, provided the vertical height is same in all cases. This is even true for curved path. Any curved path can be considered to be series of linear path having horizontal and vertical components. As horizontal component is not relevant, vertical components sum up to the net height.
We must, however, emphasize that this is not a general result, but specific to gravity and other conservative forces (we shall define this concept in a separate module.). For example, if the surfaces are rough, then block will not be able to retain its kinetic energy on the return to initial position of projection due to friction, which is not a conservative force.
In the nutshell, we have learned following characterizing features of the motion of block acted upon by gravity alone :
We must understand that work by force due to gravity for a given displacement is independent of the presence of other forces. The work done by gravity remains same. When other forces are also present, “work-kinetic energy” theorem has following form :
$$\begin{array}{l}\Rightarrow {K}_{f}-{K}_{i}={W}_{G}+{W}_{F}\end{array}$$
where ${W}_{G}$ and ${W}_{F}$ are the work done by the gravity and other force(s).
Here, work done by other external force(s) may be analyzed with respect to following different conditions :
In the first two cases, initial and final kinetic energy are same. Hence,
$$\begin{array}{l}\Rightarrow {K}_{f}-{K}_{i}={W}_{G}+{W}_{F}=0\\ \Rightarrow {W}_{F}=-{W}_{G}\end{array}$$
This is an important result. The work done by other force(s) is same as that by gravity, but with a negative sign (we have already discussed such case in previous module). This feature of motion, when end conditions are same, facilitates computation of work by other forces a great deal, as we need not be concerned of the external force other than gravity. We simply compute the work by gravitational force. The work by other external force(s) is equal to the work by gravity with a negative sign.
In third case, kinetic energies at end points are not same. The difference in kinetic energy during a motion is equal to the net work done by all forces. In this case, we can not consider work by other forces to be equal in magnitude to that by gravity. Instead, we would require to find individual force by force analysis and then compute work by each force individually.
Problem : A lift, weighing 1000 kg, is raised vertically by a string - pulley arrangement with an upward acceleration of 1 $m/{s}^{2}$ . If the initial speed of the lift is 2 m/s, find the speed attained by the lift after traveling a vertical distance of 10 m.
Solution : Here, lift is moving with upward acceleration of 1 $m/{s}^{2}$ . It means that velocities in the beginning and end are not same. Therefore, work done by gravity and tension is not zero.
Now, work done by gravity is :
$$\begin{array}{l}\Rightarrow {W}_{G}=-mgy=-1000x10x10=-100000\phantom{\rule{2pt}{0ex}}J\end{array}$$
In order to know the work done by the tension, we need to know the tension. Note here that work by tension is not equal to work by gravity as in earlier case. From, free body diagram, we have :
$$\begin{array}{l}\Rightarrow T-mg=ma\\ \Rightarrow T=m(g+a)=1000x(10+1)=11000\phantom{\rule{2pt}{0ex}}N\end{array}$$
Work done by tension in the string is :
$$\begin{array}{l}\Rightarrow {W}_{T}=Txy\\ \Rightarrow {W}_{T}=11000x10=110000\phantom{\rule{2pt}{0ex}}J\end{array}$$
Using “work - kinetic energy” theorem, we have :
$$\begin{array}{l}{K}_{f}-{K}_{i}={W}_{G}+{W}_{F}=-100000+110000=10000\phantom{\rule{2pt}{0ex}}J\\ \Rightarrow \frac{1}{2}m({{v}_{f}}^{2}-{{v}_{i}}^{2})=10000\end{array}$$
$$\begin{array}{l}\Rightarrow {{v}_{f}}^{2}=\frac{2x10000}{m}+{{v}_{i}}^{2}\end{array}$$
$$\begin{array}{l}\Rightarrow {{v}_{f}}^{2}=20+{2}^{2}\\ \Rightarrow {{v}_{f}}^{2}=24\\ \Rightarrow {v}_{f}=4.9\phantom{\rule{2pt}{0ex}}m/s\end{array}$$
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