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

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)

Calculation:

The length l is equal to half of the circle's perimeter: l = πR. Thus, the equilibrium positions of the balls A and B are at P and Q, respectively.

Let θ be the angular displacement of each ball from its equilibrium position. In this condition, one spring is compressed by length Rθ and the other is extended by the same length. So the restoring force on ball B by the two springs is:

F = k(2Rθ) + k(2Rθ) = 4kRθ

The tangential acceleration at of the ball B is:

at = R × d²θ/dt²

Apply Newton's second law:

mR × d²θ/dt² = -4kRθ

⇒ d²θ/dt² = -(4k/m)θ

This is the equation of SHM with angular frequency ω² = 4k/m

Frequency, ν = (1 / 2π) × √(4k / m) = (1 / 2π) × √(0.4 / 0.1) = 1 / π Hz

At the initial position (θ = π/6), both springs are stretched/compressed by Rθ. Potential energy of each spring:

U1,i = U2,i = (1/2) × k × (2Rθ)² = 2kR²θ²

Total initial potential energy of system:

Ui = 4kR²θ²

Initial kinetic energy = 0

Final position: each spring is relaxed, so Uf = 0

Total kinetic energy at final position:

Kf = m × v² (since both balls move at same speed)

Using conservation of energy:

Ui + Ki = Uf + Kf

4kR²θ² = m × v²

Solving for v:

v = 2Rθ × √(k / m) = 2 × 0.06 × (π / 6) × √(0.1 / 0.1) = 0.0628 m/s

Total energy of the system:

E = Ui = Kf = 4kR²θ²

E = 4 × 0.1 × (0.06)² × (π / 6)² = 3.9 × 10⁻⁴ J

Thus n / α = 4/ 3.9 ≈ 1

Calculation:

The bullet of mass m = 20 g = 0.02 kg first collides with a plate of mass m1 = 1 kg and then with a second plate of mass m2 = 2.98 kg.

Let u be the initial velocity of the bullet, and v be its velocity after passing through m1. Let v1 be the velocity of m1 after the bullet passes through it.

Apply conservation of linear momentum before and after first collision:

mu = mv + m1v1      (1)

Now, the bullet embeds into the second plate m2, so they move with common velocity v1 (since all move together after collision).

Apply conservation of linear momentum again before and after second collision:

mv + m1v1 = (m + m2)v1      (2)

Now eliminate v1 from equations (1) and (2):

v / u = (m + m2) / (m + m1 + m2)

= (0.02 + 2.98) / (0.02 + 1 + 2.98) = 3.00 / 4.00 = 0.75

The bullet retains 75% of its velocity. So, percentage loss in velocity = 25%

Answer: 25%

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