Work Power and Energy MCQ Quiz - Objective Question with Answer for Work Power and Energy - Download Free PDF

Calculation:

Let mass = m, radius = r, and the object rolls without slipping.

Then, v = rω

Total energy at the start:

Kinetic translational = (1/2)mv2

Kinetic rotational = (1/2)Iω2 = (1/2)I(v2/r2)

Total = (1/2)mv2(1 + I/mr2)

Final energy at height = Potential energy = mgh = m × (3v2/2g) = (3/2)mv2

Equating energies:

(1/2)mv2(1 + I/mr2) = (3/2)mv2

→ 1 + I/mr2 = 3

→ I/mr2 = 1

→ I = mr2

This matches the moment of inertia of a solid disc.

Correct Answer: Option 2

Calculation:

Let ρ be the density of air, and A the area swept by the turbine blades.

Volume of air hitting the blades per second = A × v

Mass per second = ρ × A × v

Kinetic energy per unit mass = (1/2) v2

Power = (1/2) × ρ × A × v × v2 = (1/2) × ρ × A × v3

Hence, P ∝ v3

Correct Answer: Option C

The correct answer is 490J

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,

Mass (m) = 20 kg

Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s

According to the work-energy theorem,

⇒  Work done = Change in K.E

⇒  W = Δ K.E

Since initial speed is zero so the initial KE will also be zero.

⇒  Work done (W) = Final K.E = 1/2 mv²

⇒  W = 1/2 × 20 × 7²

⇒  W = 10 × 49

⇒  W = 490J

The correct answer is 2 kJ.

CONCEPT:

• Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.

Potential energy is given by:

PE = m g h.

Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed

CALCULATION:

Given that: 

Mass (m) = 10 Kg

Height (h) = 20 m

P.E. = 10 x 10 x 20

P.E.= 2000 J

 P.E. = 2 kJ

• Kinetic energy: The energy due to the motion of the object is called kinetic energy. 
  • Kinetic energy (KE) = 1/2 (mv²)
  • Where m is mass and v is velocity. 
• Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
• Only the potential energy of the object will be there at the height.

Calculation:

(A) Force is zero at x = -a, 0, a. Oscillation about x = ±a for energy U₀ / 4 → (p), (q), (r), (t)

(B) Force is zero at x = 0. It's attractive near x = 0 → (q), (s)

(C) Force is zero at x = -a, 0, a. Potential well around x = 0 → (p), (q), (r), (s)

(D) Force is zero at x = -a and x = a. Oscillation possible around x = -a for U₀ / 4 → (p), (r), (t)

The correct answer is 2 kJ.

CONCEPT:

• Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.

Potential energy is given by:

PE = m g h.

Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed

CALCULATION:

Given that: 

Mass (m) = 10 Kg

Height (h) = 20 m

P.E. = 10 x 10 x 20

P.E.= 2000 J

 P.E. = 2 kJ

• Kinetic energy: The energy due to the motion of the object is called kinetic energy. 
  • Kinetic energy (KE) = 1/2 (mv²)
  • Where m is mass and v is velocity. 
• Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
• Only the potential energy of the object will be there at the height.

The correct answer is 490J

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,

Mass (m) = 20 kg

Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s

According to the work-energy theorem,

⇒  Work done = Change in K.E

⇒  W = Δ K.E

Since initial speed is zero so the initial KE will also be zero.

⇒  Work done (W) = Final K.E = 1/2 mv²

⇒  W = 1/2 × 20 × 7²

⇒  W = 10 × 49

⇒  W = 490J

CONCEPT:

• Kinetic energy: The energy needed to move the body of mass m from one point to another with stated velocity v is called kinetic energy.

The Kinetic energy is given as:

K.E = ½ × m V²

Where K.E = Kinetic Energy

m = mass of the object

V = Velocity of an object

CALCULATION:

Given that, m = 22 kg, v = 5 m/s

∴ K.E = ½ × 22 × 5²

K.E = 275 J

CONCEPT:

• Kinetic energy (KE): The energy possessed by a body by virtue of its motion is called kinetic energy.

\(KE = \frac{1}{2}m{v^2}\)

Where m = mass of the body and v = velocity of the body

CALCULATION:

Given that:

Initial speed (u) = 10 m/s

Final speed (v) = 20 m/s

Let the total mass is m.

\(KE = \frac{1}{2}m{v^2}\)

The ratio of final KE to initial KE is given by:

\(Ratio = \frac{\frac{1}{2}mv^2}{\frac{1}{2}mu^2} = \frac{v^2}{u^2} =\frac{20^2}{10^2}=4:1\)

Concept:

• Potential energy is defined as the energy stored due to change in position relative to others, stresses within itself, or many factors.
  • The potential energy (U) = m g h [where m= mass of body, g= acceleration due to gravity, h = distance from the ground].
  • If the height of a body increases from the ground its energy also increases and vice versa.

Calculation:

Work-done to carry a box up to height = m g h

Here the person himself do work on his own mass + mass of box

Let the mass of the person be x

So,

Work-done by potential energy = mass (both) × g × height

⇒ 5000 J = (x kg + 15 kg) × 10 × 5

⇒ 5000 = (x+15) × 50

⇒ 100 = x + 15

⇒ x = 85 kg

Hence the mass of the person is 85 kg.

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,
Initial velocity (u) = 30 km/h = (30 × 1000/3600) = 25/3 m/s

Initial Kinetic energy (KE_i) = 52000 = \(\frac{1}{2}mu^2\)

Final Velocity (v) = 60 km/h = (60 × 1000/3600) = 50/3 m/s = 2u

Final kinetic energy (KE_f) = \(\frac{1}{2}mv^2 = \frac{1}{2}m (2u)^2 =4 KE_i \)

⇒ KE_f  = 4 × 52000 = 208000 J

• According to the work-energy theorem, 

⇒  Work done = Change in K.E
⇒  Work done (W) = Δ K.E = KE_f - KE_i = 208000 - 52000 = 156000 J

CONCEPT:

• Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
• Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:

\(W = \vec F \cdot \vec s\)

Or, W = Fs cos θ

• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
  • Work done is positive when the direction of both force and displacement are the same
  • Work done is negative when the direction of force and displacement are the opposite.
  • Whereas work is done is zero when direction force (Centripetal force) and displacement are perpendiculars this case is true for an object moving in a circular motion

CALCULATION:

Given that:

Mass of the car (m) = 1000 kg

Velocity (v) = 15 m/s 

• The car is moving on the horizontal plane, so the displacement is in a horizontal direction.
• The gravity force (equal to the weight of the car) is acting in a vertically downward direction.
• So the angle between the force (weight) and the displacement is 90°.

Work done (W) = Fs Cos90° = 0

So option 1 is correct.

CONCEPT:

• Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
• Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by:

\(W = \vec F \cdot \vec s\)

Or, W = Fs cos θ

• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.

CALCULATION:

Given that:

Force (F) = 12 N

Displacement (s) = 60 cm = 60/100 = 0.6 m

The force (F) and displacement (s) are in the same direction, so θ = 0° 

Work done = Fs cos θ = 12 × 0.6 × 1 = 7.2 J

Hence option 2 is correct.

CONCEPT:

Power:

• The rate of doing work is called power.
• SI unit of power is the watt.

\(⇒ P=\frac{W}{t}\)

.Where P = power, W = work done and t = time

CALCULATION:

Given m = 20 kg, h = 5 m, t = 20 sec and g = 10 m/sec²

• We know that the work done in lifting a body of mass m to height h is given as,

⇒  W = mgh    -----(1)

By equation 1,

⇒  W = 20 × 10 × 5 = 1000 J 

• So power required,

\(⇒ P=\frac{W}{t}\)

\(⇒ P=\frac{1000}{20}\)

\(⇒ P=50\,watt\)

• Hence, option 1 is correct.

CONCEPT:

• Work is said to be done by a force when the body is displaced actually through some distance in the direction of the applied force.
• Since the body is being displaced in the direction of F, therefore work done by the force in displacing the body through a distance s is given by

\(W = \vec F \cdot \vec s\)

Or, W = Fs cos θ

• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.

CALCULATION:

Given that:  

Force (F) = 50 , Displacement (s) = 5 m and  θ = 0 (because the force is in the direction of motion)

To find the amount of work done in this case, we must apply the following formula:

⇒ W = Fs cosθ

⇒ W = 50 × 5 = 250 J