Question

A particles with a change of 2.5 columb is take from a point A at a potential of 60v to another point B at a potential of 130v calculate work done

Answer 1

Q=2.5C, V1 =60v, V2 =130v
V2 - V1 = W/Q
130 - 60 = W/2.5
70 = W/2.5
W = 70 * 2.5
W = 175J

Answer 2

 Work done against gravity - gravitational potential energy

So far we have only considered objects falling under gravity. Let's now consider the work done when we lift an object. In order to lift an object that has mass m, we have to apply an upward force mg to overcome the downward force of gravity. If this force raises the object through a height h, then the work done is:

Figure 5 (a) Placing a suitcase on a luggage rack involves doing work against gravity. (b) The stored energy is released if the suitcase falls off the rack.

So if an object of mass m is raised through a height h, the work done on the object is equal to mgh, and so this amount of energy is transferred to the object. (Notice that this equation is identical to the one describing an object falling under gravity, Equation 7.)

Of course, this ties in very well with everyday observations. If you lift a heavy suitcase onto a luggage rack in a train, or a heavy bag of shopping onto a table, you are very aware that you are doing work against gravity. You will also be aware that more work is required to lift a more massive object, or the same object to a greater height, and these 'observations' are consistent with the work done being equal to mgh.