Learning Objectives:

• Calculate the change in kinetic energy or speed that results from performing a specified amount of work on an object.

• Apply the theorem to determine the change in an object’s kinetic energy and speed that results from the application of specified forces, or to determine the force that is required in order to bring an object to rest in a specified distance.

Kinetic Energy of an object is the energy that it possesses due to its motion. The kinetic energy of a non-rotating object of mass m traveling at a speed v is:

\[KE = \dfrac{1}{2} mv^2\]

Solved Example: 31-1-01

A body possesses linear momentum 2 kg.m/s and kinetic energy of 2J. The mass of the body is:

A. 0.5 kg

B. 1.42 kg

C. 1 kg

D. 2 kg

Linear momentum, \[mv = 2\] Also, kinetic energy, \[\dfrac{1}{2} mv^2 = 2\] Dividing these two equations, we get \[\dfrac{2}{v} = 1\] \[v = 2\] Substituting in the first equation, m = 1 kg

Correct Answer: C

Solved Example: 31-1-02

The kinetic energy of a particle having mass m is E. Its momentum will be: (UPTET 2019 Paper II- Mathematics and Science)

A. 2 mE

B. $\sqrt{\dfrac{1}{2} mE}$

C. $\sqrt{2 mE}$

D. $\dfrac{1}{2} mE$

\[KE = \dfrac{1}{2}mv^2\] Given KE = E, \begin{align*} E &= \dfrac{1}{2}mv^2\\ 2E &= mv^2\\ \dfrac{2E}{m} &= v^2\\ \sqrt{\dfrac{2E}{m}} &= v\\ \end{align*} Momentum is given by, \begin{align*} M &= mv\\ &= m \times \sqrt{\dfrac{2E}{m}}\\ &= \sqrt{m} \sqrt{m} \times \sqrt{\dfrac{2E}{m}}\\ &= \sqrt{m} \times \sqrt{2E}\\ &= \sqrt{2mE}\\ \end{align*}

Correct Answer: C

Solved Example: 31-1-03

During an object's free fall, which of the following energies increases at one point in its path? (RRC Group D Sep 2018 Shift I)

A. Kinetic energy

B. Chemical energy

C. Potential energy

D. Mechanical energy

Correct Answer: A

Solved Example: 31-2-04

What type of energy conversion takes place during the thunder of clouds? (RRC Group D Nov 2018 Shift III)

A. Frictional energy is converted into light and sound energy.

B. Kinetic energy is converted into light and sound energy.

C. Potential energy is converted into light and sound energy.

D. Kinetic energy is converted into electrostatic energy.

Correct Answer: B

Learning Objectives:

• Calculate the gravitational potential energy of a particle in various situations.

• Calculate the elastic potential energy of a particle attached to a stretched spring.

Potential Energy is the energy stored in an object due to its position.

Gravitational Potential Energy

\[U = mgh\]

Elastic Potential Energy

\[U = \dfrac{1}{2} k x^2\]

where k = stiffness(spring) constant

Solved Example: 31-2-01

Jack climbs a mountain via a short, steep trail. Jill ascends the same mountain via a long and considerably less steep path. Which of the following statements is true? Assume that Jack and Jill have the same mass.

A. Jack gains more gravitational potential energy than Jill.

B. Jack gains less gravitational potential energy than Jill.

C. Jack gains the same gravitational potential energy as Jill.

D. To compare gravitational potential energies, we must know the height of the mountain.

$\Delta U = mgh$, same h for both. So gravitational potential energy is same.

Reference: http://resources.seattlecentral.edu

Correct Answer: C

Solved Example: 31-2-02

A compressed spring possesses _________. (RRC Group D Sept 2018 Shift II)

A. Electric energy

B. Chemical energy

C. Potential energy

D. Kinetic energy

Correct Answer: C

Solved Example: 31-2-03

A ball is dropped from a height h and rebounds to a height which is 80 % of the initial height. Find the ratio of final potential energy to the initial potential energy of the ball. (Airforce Group X Nov 2020)

A. $\dfrac{5}{4}$

B. $\dfrac{4}{5}$

C. $\dfrac{25}{4}$

D. $\dfrac{4}{25}$

Correct Answer: B

Solved Example: 31-2-04

The energy possessed by a body by virtue of its position is called the _________.(HSSC Lower Division Clerk Feb 2020 Shift III)

A. Electrostatic Energy

B. Kinetic Energy

C. Conservative Energy

D. Potential Energy

Correct Answer: D

Learning Objectives:

• Calculate the work done by a specified constant force on an object that undergoes a specified displacement.

• Relate the work done by a force to the area under a graph of force as a function of position, and calculate this work in the case where the force is a linear function of position.

• Calculate the work performed by a force that supplies constant power, or the average power supplied by a force that performs a specified amount of work.

A force is said to do work when acting on a body there is a displacement of the point of application in the direction of the force. The SI unit of work is the Joule (J), which is defined as the work expended by a force of one newton through a distance of one meter. \[W = F \times s\] where,
W = Work in Joules
F = Force in Newtons
s = displacement (shortest distance between two points), in meters Work U is defined as \[\int \bar{F} \cdot \bar{dr}\] Variable force \[\int F \cos \theta ds\] Constant force \[F_c \cos \theta \Delta s\] Weight \[U = -W \Delta y\] Spring \[U_s = -(1/2 ks_2^2 - 1/2 ks_1^2)\] Couple moment \[U = M \Delta \theta\]

Solved Example: 31-3-01

A truck driver slams on the brakes of a moving truck with a constant velocity v, as a result of his action the truck stops after traveling a distance d. If the driver had been traveling with twice the velocity, what would be the stopping distance compared to the distance in the first trial?

A. Two times greater

B. Four times greater

C. The same

D. Half as much

The acceleration (in this case retardation) depends upon Force and mass, both of which are unchanged in two cases. Hence $a= \dfrac{F}{m}$ will remain unchanged irrespective of the velocity.

Here the final velocity(v) is zero, initial velocity (u) is $V_{initial}$
Using, \begin{align*} v^2 &= u^2 + 2as\\ 0 &= u^2 + 2as\\ u^2 &= -2as\\ s &= -\dfrac{u^2}{2a} \end{align*} Hence, the stopping distance is proportional to the square of initial velocity, provided other parameters remain same. That means, if truck speed is doubled, the stopping distance will now be four times greater.

Correct Answer: B

Solved Example: 31-3-02

The work done by a person in carrying a box of mass 15 kg through a vertical height of 5 m is 5000 J. The mass of the person is _______ Take g = 10 m/s$^2$. (DSSSB TGT Natural Science Sept 2018 Shift II)

A. 95 kg

B. 45 kg

C. 65 kg

D. 85 kg

Let 'x' be the mass of the person. Hence, the combined mass (person + box) raised will be (x+ 15)kg. \begin{align*} W &= mgh\\ 5000 &= (x + 15) \times 10 \times 5\\ x + 15 &= 100\\ x &= 85\ kg \end{align*}

Correct Answer: D

Learning Objectives:

• Define Shaft Power, fluid power, volumetric efficiency and overall efficiency.

• Calculate values in a fluid power system utilizing Pascal’s law.

• Calculate flow rate, flow velocity and mechanical advantage in a hydraulic system.

• Learn and understand principles of fluid mechanics and power with applications.

• Provide the student the necessary analytic skills to solve and analyze a variety of fluid mechanics and fluid power related problems.

Mechanical Power is the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the MKS system, the unit of power is the joule per second (J/s), known as the Watt in honor of James Watt. Larger units are 1KW = 1000 W or 1 Horsepower (HP) = 746 W.
Electric Power in watts produced by an electric current I consisting of a charge of Q coulombs every t seconds passing through an electric potential (voltage) difference of V is: \[P = V \times I = \dfrac{V^2}{R} = I^2 R\] Where,
P = Electric power in Watts
t = time in seconds
I = electric current in Amperes
V = electric potential or voltage in Volts
Mechanical Power, \(P = \dfrac{dU}{dt} = \bar{F} \cdot \bar{v}\)
Mechanical Efficiency, \(\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{U_{out}}{U_{in}}\)

Solved Example: 31-4-01

Amount of energy supplied by current in unit time is called:

A. Electric energy

B. Friction

C. Resistance

D. Electric power

Power is rate of work done (or energy supplied). \[\mathrm{Power} = \dfrac{\mathrm{Energy}}{\mathrm{time}}\] \[P = \dfrac{E}{t}\]

Correct Answer: D

Solved Example: 31-4-02

A 220 $\Omega$ resistor dissipates 3 W. The voltage is:

A. 73.3 V

B. 2.5 V

C. 25.7 V

D. 257 V

Using, \[P = VI = I^2R = \dfrac{V^2}{R}\] Substituting the values, \begin{align*} P &= \dfrac{V^2}{R}\\ 3 &= \dfrac{V^2}{220}\\ V^2 &= 220 \times 3\\ V &= 25.7\ V \end{align*}

Correct Answer: C

Solved Example: 31-4-03

A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in m/s) of the particle is zero, the speed (in m/s) after 5 s is:

A. 2.0

B. 2.5

C. 5.0

D. 5.6

Power is rate of Energy. \begin{align*} P &=\dfrac {dE}{dt}\\ P &=\dfrac {d}{dt}\left( \dfrac {1}{2}mv^{2}\right) \\ P &=\dfrac {1}{2}m\dfrac {d}{dt}\left( v^{2}\right) \\ \dfrac {2P}{m}&=2v\dfrac {dv}{dt}\\ \dfrac {P}{m}dt&=vdv\\ \int ^{t}_{0}\dfrac {P}{m}dt&=\int ^{v}_{0}vdv\\ \dfrac {P}{m}t&=\dfrac {v^{2}}{2}\\ v^{2}&=\dfrac {2\times 0.5\times 5}{0.2}\\ v^{2}&=25\\ v&=5\ m/s \end{align*}

Correct Answer: C

Learning Objectives:

• Interpret (and provide) examples that illustrate the law of conservation of energy such as transformation from potential and kinetic energy, and vice versa.

If T\(_i\) and V\(_i\) are, respectively, the kinetic and potential energy of a particle at state i, then for conservative systems (no energy dissipation or gain), the law of conservation of energy is \[T_2 + V_2 = T_1 + V_1\] If nonconservative forces are present, then the work done by these forces must be accounted for. Hence \[T_2 + V_2 = T_1 + V_1 + U_{1 \rightarrow 2}\] where U\(_{1 \rightarrow 2}\) = the work done by the nonconservative forces in moving between state 1 and state 2.
Care must be exercised during computations to correctly compute the algebraic sign of the work term.

• If the forces serve to increase the energy of the system, U\(_{1 \rightarrow 2}\) is positive.

• If the forces, such as friction, serve to dissipate energy, U\(_{1 \rightarrow 2}\) is negative.

Solved Example: 31-5-01

Energy can neither be created nor be destroyed, but it can be changed from one form to another", this law is known as:

A. Kinetic energy

B. Potential energy

C. Conservation of energy

D. Conservation principle

Correct Answer: C

Solved Example: 31-5-02

What cannot happen to energy according to the law of conservation of energy?

A. Energy cannot be transformed.

B. Energy cannot be created

C. Energy cannot be changed.

D. Energy cannot be converted.

Correct Answer: B

Solved Example: 31-5-03

Energy cannot be created nor destroyed, only changed from one form to another. How does listening to music on a radio obey the law of conservation of energy?

A. Sound energy is changed into potential energy.

B. Electrical energy is gradually destroyed as the radio plays.

C. Electrical energy remains unchanged.

D. Electrical energy is converted into other forms of energy, such as sound.

Correct Answer: D