Work is said to be done when applied force is able to produce some displacement.It is treated like a dot product of force and displacement and it is a scalar. Work is said to be done when we have energy and it is measured in the unit called Joule. We shall multiply the displacement with the component of the displacement acting along the direction of the displacement.Work can be studied by taking only its magnitude into consideration and it don't need direction to understand it. Thus it is treated like a scalar.
Problem
It is given in the problem that a force in vector form is acting on a particle and the particle got displacement both along X and Y axis and its values are given to us. We need to know the total work done in this case.
Solution
We know that the work done is the dot product of force and displacement. We know that the dot product of unit vectors along the same direction is one and the dot product of unit vectors along perpendicular directions is zero. Taking that into consideration, we can solve the problem as shown in the diagram below.
Problem
Water is lifted from a well of the depth 50 meter and the rope used to lift is also having a mass of 0.2 kilogram per meter and mass lifted is 20 kilogram, we need to measure the work done in this case.
Solution
We know that the work done in this case is stored in the form of potential energy and we need to measure it. Generally mass of the rope is small and it is neglected. But in this problem mass of the rope per kilogram is given to us. To lift the water we need a rope equal to the depth of the water and we need to consider the mass of the entire rope and measure the work done in lifting the water as well as rope.
Problem
It is given in the problem that a meter stick of mass 400 gram is lying on the horizontal surface and we need to get into vertical position. We need to know the work done in this process.
Solution
When the scale is in the horizontal position and it has no potential energy as it is the reference point from which we measure the potential energy. As we lift it vertically, some work is done and it is in the form of potential energy. We shall not consider the height of the scale as the entire length of the scale. Its center of mass is at the geometrical center of the scale and we need to measure that only the height to which the body is lifted. Center of mass is a point of the system where the entire mass appears to be concentrated. We can solve the problem as shown in the diagram below.
Problem
A rain drop of known radius is falling from a certain height and we need to know the work done relation with the radius of the drop. The problem is as shown in the diagram below.
Solution
We know that the drop is falling from a certain height and the energy is stored in the form of potential energy. We know that the potential energy is directly proportional to mass of drop and it can be expressed as the product of volume and density. As the drop is in the spherical format, volume is directly proportional to the cube of the radius.
Problem
A can of water 10 kilogram is raised from a depth of 20 meter. When the can rises by 10 meter it has a known velocity. We need to measure the work done in this process. The problem is as shown below.
Solution
The body is not only taken to a height and it is also having some velocity at that instant. Thus the body has both potential and kinetic energies at the given instant. Thus the body has both the energies and the work done is sum of both the energies. The problem can be solved as shown below.
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