When an object is thrown upwards its kinetic energy goes on decreasing and consequently its potential energy keeps increasing (neglecting air resistance).

Law of conservation of energy

When an object is thrown upwards its kinetic energy goes on decreasing and consequently its potential energy keeps increasing (neglecting air resistance). When it reaches the highest point its energy is completely potential. Similarly, when the object falls back from a height its kinetic energy increases whereas its potential energy decreases. When it touches the ground its energy is completely kinetic. At the intermediate points the energy is both kinetic and potential as shown in Figure 4.13. When the body reaches the ground the kinetic energy is completely dissipated into some other form of energy like sound, heat, light and deformation of the body etc.

In this example the energy transformation takes place at every point. The sum of kinetic energy and potential energy i.e., the total mechanical energy always remains constant, implying that the total energy is conserved. This is stated as the law of conservation of energy.

The law of conservation of energy states that *energy can neither be created nor* *destroyed. It may be transformed from one form to another but the total energy of an isolated system remains constant.*

Figure 4.13 illustrates that, if an object starts from rest at height *h*, the total energy is purely potential energy (U=mgh) and the kinetic energy (KE) is zero at *h*. When the object falls at some distance *y*, the potential energy and the kinetic energy are not zero whereas, the total energy remains same as measured at height *h*. When the object is about to touch the ground, the potential energy is zero and total energy is purely kinetic.

An object of mass 1 kg is falling from the height *h* = 10 m. Calculate

a. The total energy of an object at *h* =10 m

b. Potential energy of the object when it is at *h* = 4 m

c. Kinetic energy of the object when it is at *h* = 4 m

d. What will be the speed of the object when it hits the ground?

(Assume *g* =10 m s-2)

a. The gravitational force is a conservative force. So the total energy remains constant throughout the motion. At *h *=10* *m, the total energy*E *is entirely* *potential energy.

b. The potential energy of the object at *h *=* *4* *m is

c. Since the total energy is constant throughout the motion, the kinetic energy at *h* = 4 m must be *KE *=* E *-* U *=* *100* *-* *40* *=* *60J

Alternatively, the kinetic energy could also be found from velocity of the object at 4 m. At the height 4 m, the object has fallen through a height of 6 m.

The velocity after falling 6 m is calculated from the equation of motion,

d. When the object is just about to hit the ground, the total energy is completely kinetic and the potential energy, *U* = 0.

A body of mass 100 kg is lifted to a height 10 m from the ground in two different ways as shown in the figure. What is the work done by the gravity in both the cases? Why is it easier to take the object through a ramp?

*Solution*

m = 100 kg, h = 10 m

*Along path (1):*

The minimum force *F*1 required to move the object to the height of 10 m should be equal to the gravitational force, F1 mg = 100 x 10 = 1000 N

The distance moved along path (1) is, =10 m

The work done on the object along path (1) is

W = Fh = 1000 x 10 = 10,000 J

*Along path (2):*

In the case of the ramp, the minimum force *F*2* *that we apply on the object to take it* *up is not equal to *mg*, it is rather equal to *mg *sin*θ** *.*(*mg sin <* mg) *.

Here, angle θ = 30o

Therefore, F2 = mg sinθ = 100 × 10 × sin30o = 100 × 10 × 0.5 = 500N

Hence, (mg sinθ < mg)

The path covered along the ramp is,

l = h/sin30 = 10/0.5 =20m

The work done on the object along path (2) is, W = F2 l = 500 × 20 = 10,000 J

Since the gravitational force is a conservative force, the work done by gravity on the object is independent of the path taken.

In both the paths the work done by the gravitational force is 10,000 J

Along path (1): more force needs to be applied against gravity to cover lesser distance .

Along path (2): lesser force needs to be applied against the gravity to cover more distance.

As the force needs to be applied along the ramp is less, it is easier to move the object along the ramp.

An object of mass m is projected from the ground with initial speed v0.

Find the speed at height h.

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

Final values of potential energy, kinetic energy and total energy are measured at the height *h*.

By law of conservation of energy, the initial and final total energies are the same.

Note that in section (2.11.2) similar result is obtained using kinematic equation based on calculus method. However, calculation through energy conservation method is much easier than calculus method.

An object of mass 2 kg attached to a spring is moved to a distance *x* =10 m from its equilibrium position. The spring constant *k *=1* *N m-1and assume that the surface is* *frictionless.

a. When the mass crosses the equilibrium position, what is the speed of the mass?

b. What is the force that acts on the object when the mass crosses the equilibrium position and extremum position *x* = ± 10 m.

a. Since the spring force is a conservative force, the total energy is constant. At x =10 m, the total energy is purely potential.

When the mass crosses the equilibrium position *x* = 0 , the potential energy

The entire energy is purely kinetic energy at this position.

b. Since the restoring spring force is F = - kx, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at *x* = +10 m (elongation), the force F = - k *x*

F = - (1) (10) = - 10 N. Here the negative sign implies that the force is towards equilibrium i.e., towards negative *x*-axis and when the object is at *x* = - 10 (compression), it experiences a forces F = - (1) (- 10) = +10 N. Here the positive sign implies that the force points towards positive *x*-axis.

The object comes to momentary rest at *x* = ±10*m* even though it experiences a maximum force at both these points.

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11th Physics : UNIT 4 : Work, Energy and Power : Law of conservation of energy |

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