Question
Question: A stone of mass 6kg is revolved in a vertical circle of diameter 6m, such that its speed at a point ...
A stone of mass 6kg is revolved in a vertical circle of diameter 6m, such that its speed at a point is minimum. If the K.E at the same point is 250J, then the P.E. at this point is
A. 200J
B. 150J
C. 100J
D. 450J
Solution
The minimum velocity required for a particle to loop a loop while going in a vertical circle of radius r is 5gr. At the base, the P.E. of a body is zero and there will be only K.E. Use the law of conservation of mass to find the P.E. of the body at a point where the speed is minimum.
Complete step by step answer:
We are given that a stone of mass 6kg is revolved in a vertical circle of diameter 6m, such that its speed at a point is minimum. The point is P from the above figure. The K.E. at point P is 250J.
We have to find the Potential Energy at point P.
Q is the starting point for the stone. At Q, the potential energy is zero and only kinetic energy will be there.
The velocity of the stone at point Q is 5gr, where r is the radius and g is the acceleration due to gravity.
r=2d d=6m ⇒r=26=3m r=3m,g=10m/s2
⇒v=5gr ⇒v=5×10×3 ⇒v=150=56m/s
By Law of conservation of Energy, total mechanical energy is equal to sum of kinetic energy and potential energy. The Mechanical energy at point P and Q must be equal as the energy can be neither created nor destroyed but can only be transformed.
M.Ep=M.Eq ⇒K.Ep+P.Ep=K.Eq+P.Eq P.Eq=0 K.Ep=250J ⇒K.Eq+0=250J+P.Ep K.Eq=21mv2 v=56m/s ⇒K.Eq=21×6×(56)2 ⇒K.Eq=3×150=450J ⇒450J=250J+P.Ep ⇒P.Ep=450−250=200J
The potential energy at a point where the speed is minimum is 200J.
The correct answer is Option A.
Note: Kinetic energy of a system is the energy acquired by it by virtue of its motion whereas potential energy of a system is the energy acquired by it by virtue of its position. They are not the same. So do not confuse Kinetic Energy with Potential energy.