Law of conservation of mechanical energy l my experiences

Describe the law of conservation of mechanical energy?

Law of conservation of mechanical energy - The sum of object’s kinetic energy and potential energy is called mechanical energy. If a body is under the action of conservative force/forces alone, the total mechanical energy of the body remains constant. This is called law of conservation of mechanical energy. The name “conservative” force is appropriate because a conservative force corresponds to the conservation of constancy of mechanical energy. The law of conservation of mechanical energy is best illustrated by example like a free falling body or a body sliding down a smooth inclined plane, its K.E. is starts increasing. The gain in K.E. is at the expense of loss in gravitational potential energy but total mechanical energy of the body remains conserved. We shall discuss below the two examples.

A) Freely falling body

i ) Consider a body of mass m at rest at point A at the height h above the ground as the following figure shows. Let the body is dropped: As the body falls freely under the gravity i.e., with acceleration g, its height and hence P.E. goes on decreasing, but its velocity and hence its K.E. goes on increasing. The decrease in P.E at any instant is equal to gain in K.E., so that total energy is constant. This could be understood as follows.
ii) At the position A
Potential energy of the body = mgh
Kinetic energy of the body = ½ mv_A² =0;
(as v_A = 0 at A)
Total mechanical energy of the body at A= K.E + P.E =0 + mgh = mgh
iii) At the position B
Let the body be at the position B at any instant after having fallen through distance x. Because the body is at a height (h – x) above the ground, hence potential energy of the body at point B, then
v_B² – 0 = 2gx or v_B² = 2gx
Kinetic energy of the body = ½ mv_B² = ½ m x 2gx = mgx
Total mechanical energy of the body at B
= K.E. + P.E. = mgx + mg(h-x) = mgh ….(2)
iv) At the point C
When the body is at the position C, its height above the ground may be regarded as zero.
Potential energy of the body = 0
If v_c be the velocity with which the body just touches the ground, then v_c² – 0 =2gh or v_c² = 2gh
Kinetic energy of the body = ½ mv_c² = ½ m × 2gh =mgh
Total mechanical energy of the body at C = K.E. + P.E. = mgh + 0 = mgh ….(3)
v) It is clear from equation (1), (2) and (3) that total mechanical energy of the body at A, B and C (also at any other point) is the same. Thus we find total mechanical energy of the body remains conserved throughout its free fall. As the body falls down, the potential energy goes on decreasing whereas the kinetic energy goes on increasing.
vi) In the following figure shows the variation of kinetic energy, potential energy and total energy of the body with respect to its height from the ground. It is clear from the figure, the potential energy decreases linearly with decrease in height. On the other hand, the kinetic energy increases with decrease in height. Total mechanical energy is represented by a thick line parallel to the horizontal axis. This demonstrates that total mechanical energy is conserved.

(B) A body sliding down a smooth inclined plane

(i) Consider a smooth inclined plane of length l and height h, heaving angle of inclination θ. Let a body of mass m be at rest at the top A of the inclined plane.
Potential energy of the body at A = mgh
Kinetic energy of the body = 0
Total energy of the body = mgh …….(1)
(ii) If v_c be the velocity acquired by the body in sliding through a distance x, then
v_x² -0 =2ax
But acceleration, a = g sin θ
v_x² = 2gx sin θ = 2gx . (h₁/x) =2gh₁
Kinetic energy = 1/2 mv_x² =1/2m(2gh₁)=mgh₁
Potential energy = mgh₂
Total energy = mgh₁ + mgh₂ = mg(h₁+h₂)=mgh………(2)
(iii) If ve be the velocity acquired by the body at the bottom B of the inclined plane, then
v_e² -0=2(g sin θ) l = 2gl sin θ = 2gl(h/l)= 2gh
Kinetic energy at B = ½ mv_e² = ½ m(2gh) = mgh
Potential energy at B = 0
Total energy = mgh………….(3)
From the equations (10, (2) and (3), it is clear that the total mechanical energy is conserved.