Solved Example Problems for Energy (2023)

Physics : Work, Energy and Power : Solved Example Problems for Energy

Solved Example Problems forKinetic energy

Example 4.7

Two objects of masses 2 kg and 4 kg are moving with the same momentum of 20 kg m s-1.

a. Will they have same kinetic energy?

b. Will they have same speed?

Solution

a. The kinetic energy of the mass is given by

Note thatKE1≠KE2i.e., even though both are having the same momentum, the kinetic energy of both masses is not the same. The kinetic energy of the heavier object has lesser kinetic energy than smaller mass. It is because the kinetic energy is inversely proportional to the mass (KE∝1/m) for a given momentum.

b. As the momentum,p=mv, the two objects will not have same speed.

Solved Example Problems forPotential Energy

Example 4.8

An object of mass 2 kg is taken to a height 5 m from the groundg =10ms-2.

a. Calculate the potential energy stored in the object.

b. Where does this potential energy come from?

c. What external force must act to bring the mass to that height?

d. What is the net force that acts on the object while the object is taken to the height ‘h’?

Solution

a. Thepotential energyU=m g h= 2×10×5 = 100 J

Here the positive sign implies that the energy is stored on the mass.

b. This potential energy is transferred from external agency which applies the force on the mass.

d. From the definition of potential energy, the object must be moved at constant velocity. So the net force acting on the object is zero.

Solved Example Problems forElastic Potential Energy

Example 4.9

Let the two springs A and B be such that kA>kB. On which spring will more work has to be done if they are stretched by the same force?

Solution

The work done on the springs are stored as potential energy in the springs.

kA>kBimplies that UB>UA. Thus, more work is done on B than A.

Example 4.10

A body of mass m is attached to the spring which is elongated to 25 cm by an applied force from its equilibrium position.

a. Calculate the potential energy stored in the spring-mass system?

b. What is the work done by the spring force in this elongation?

c. Suppose the spring is compressed to the same 25 cm, calculate the potential energy stored and also the work done by the spring force during compression. (The spring constant, k = 0.1 N m-1).

Solution

The spring constant, k = 0.1 N m-1

Solved Example Problems forConservative and nonconservative forces

Example 4.11

Compute the work done by the gravitational force for the following cases

Solution

(As the displacement is in two dimension; unit vectors and are used)

a. Since the motion is only vertical, horizontal displacement component dxis zero. Hence, work done by the force along path 1 (of distance h).

Therefore, the total work done by the force along the path 2 is

Note that the work done by the conservative force is independent of the path.

Example 4.12

Consider an object of mass 2 kg moved by an external force 20 N in a surface having coefficient of kinetic friction 0.9 to a distance 10 m. What is the work done by the external force and kinetic friction ? Comment on the result. (Assume g = 10 ms-2)

Solution

m= 2 kg,d= 10 m,Fext= 20 N,k= 0.9.When an object is in motion on the horizontal surface, it experiences two forces.

a. External force,Fext=20 N

b. Kinetic friction,

fk=μkmg= 0.9x(2)x10=18N.

The work done by the external forceWext= Fs = 20x20 =200J

The work done by the force of kineticfrictionWk=fkd= (-18) x10=-180JHere the negative sign implies that theforce of kinetic friction is pposite to thedirection of displacement.

The total work done on the object

Wtotal= Wext + Wk= 200 J – 180 J = 20 J .

Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

Solved Example Problems forLaw of conservation of energy

Example 4.13

An object of mass 1 kg is falling from the heighth= 10 m. Calculate

a. The total energy of an object ath=10 m

b. Potential energy of the object when it is ath=4 m

c. Kinetic energy of the object when it is ath=4 m

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

(Assumeg=10 m s-2)

Solution

a. The gravitational force is a conservative force. So the total energy remains constant throughout the motion. Ath=10m, the total energyEis entirelypotential energy.

b. The potential energy of the object ath=4m is

c. Since the total energy is constant throughout the motion, the kinetic energy ath=4 m must beKE=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.

Example 4.14

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 forceF1required to move the object to the height of 10 m should be equal to the gravitational force, F1mg = 100 x 10 = 1000 N

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

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 forceF2that we apply on the object to take itup is not equal tomg, it is rather equal tomgsinθ.(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.

Example 4.15

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

Find the speed at height h.

Solution

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 heighth.

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.

Example 4.16

An object of mass 2 kg attached to a spring is moved to a distancex=10 m from its equilibrium position. The spring constantk=1N m-1and assume that the surface isfrictionless.

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 positionx=±10 m.

Solution

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

When the mass crosses the equilibrium positionx= 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 atx= +10 m (elongation), the force F = - kx

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

The object comes to momentary rest atx=±10meven though it experiences a maximum force at both these points.

Solved Example Problems forMotion in a vertical circle

Example 4.17

Water in a bucket tied with rope is whirled around in a vertical circle of radius 0.5 m. Calculate the minimum velocity at the lowest point so that the water does not spill from it in the course of motion. (g = 10 ms-2)

Solution

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11th Physics : UNIT 4 : Work, Energy and Power : Solved Example Problems for Energy |