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

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

The correct answer is option 3) i.e. 6400 J

CONCEPT:

• Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

​Work done, \(W = Δ KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)

Where m is the mass of the object, v is the final velocity of the object and u is the initial velocity of the object. 

CALCULATION:

Given that:

Mass, m = 10 g = 0.01 kg

Initial velocity, u = 1200 m/s

Final velocity, v = 400 m/s

The loss in energy = Work done

\(⇒ Δ KE = W\)

\(⇒​​​​ W=\frac{1}{2}m(v^2 -u^2)\)

\(⇒ ​​​​W=\frac{1}{2}\times 0.01\times(1200^2 -400^2)\)

⇒ W = 6400 J

The correct answer is option 3) i.e. 6400 J

CONCEPT:

• Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

​Work done, \(W = Δ KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)

Where m is the mass of the object, v is the final velocity of the object and u is the initial velocity of the object. 

CALCULATION:

Given that:

Mass, m = 10 g = 0.01 kg

Initial velocity, u = 1200 m/s

Final velocity, v = 400 m/s

The loss in energy = Work done

\(⇒ Δ KE = W\)

\(⇒​​​​ W=\frac{1}{2}m(v^2 -u^2)\)

\(⇒ ​​​​W=\frac{1}{2}\times 0.01\times(1200^2 -400^2)\)

⇒ W = 6400 J

Concept:

Kinetic Energy of a Rolling Object:

• Total Kinetic Energy: For a rolling object, such as a solid disc, the total kinetic energy is the sum of translational and rotational kinetic energies.
• Translational Kinetic Energy (K₁): It is due to the linear motion of the center of mass. The formula for translational kinetic energy is \( K_1 = \frac{1}{2} M v^2 \), where:
  • M: Mass of the object (SI unit: kg)
  • v: Velocity of the center of mass (SI unit: m/s)
• Rotational Kinetic Energy (K₂): It is due to the rotation of the object about its axis.
• The formula is \( K_2 = \frac{1}{2} I \omega^2 \), where:
  • I: Moment of inertia of the object (for a solid disc, \( I = \frac{1}{2} M R^2 \))
  • ω: Angular velocity of the object, related to linear velocity by \( \omega = \frac{v}{R} \)
• Work-Energy Theorem: The work done to stop an object is equal to its total kinetic energy. Thus, the work required to stop the rolling disc is the sum of its translational and rotational kinetic energies.

Calculation:

Given,

Mass of the disc, M = 50 kg

Velocity of the center of mass, v = 0.4 m/s

⇒ K₁ = \((\frac{1}{2} M v^2 = \frac{1}{2} \times 50 \times (0.4)^2\)

⇒ K₁ = 4 J

Moment of inertia of the solid disc, I = \(\frac{1}{2} M R^2\), and ω = \(\frac{v}{R}\)

⇒ K₂ = \((\frac{1}{2} \times \frac{1}{2} M v^2 = \frac{1}{4} \times 50 \times (0.4)^2\)

⇒ K₂ = 2 J

⇒ K_total = K₁ + K₂ = 4 + 2 = 6 J

∴ The absolute value of work done to stop the disc is 6 J.

Concept:

Kinetic Energy 

• The ability of an object to do work due to its motion is called kinetic energy.
• K.E. = 1/2mv² (where m = mass of an object and v = velocity)

Explanation:

Let mass and velocity of the heavier body be m₁ and v₁

And mass and velocity of a lighter body are m₂ and v₂

Since K.E. of the heavier and lighter bodies are equal. 

So that, = ½m₁v₁² = ½m₂v₂²

= m₁/m₂= (v₂/v₁)²

So, m₁ > m₂

m₁/m₂ >1 

So, that v₂²/v₁²>1

v22>v12

v₂>v₁

So the result shows that the velocity of lighter body v₂ is greater than heavier body v₁.

Therefore, option 3 is correct.

Additional InformationPotential Energy

• The ability of an object to do work due to the position of an object is called potential energy. 
• PE = mgh (where m = mass of an object, g = gravitational acceleration and h = heigh)
• Water collected by building a dam has potential energy.

Relation between Kinetic Energy and Momentum

• K.E = P²/2m

Where, P = Momentum (P = mv)

Important Points Momentum

• The product of the velocity and mass of an object is called momentum.
• The SI unit of momentum is kg-m/s.
• Dimension = MLT^-1

Concept:

Kinetic Energy 

• The ability of an object to do work due to its motion is called kinetic energy.
• K.E. = 1/2mv² (where m = mass of an object and v = velocity)

Explanation:

Let mass and velocity of the heavier body be m₁ and v₁

And mass and velocity of a lighter body are m₂ and v₂

Since K.E. of the heavier and lighter bodies are equal. 

So that, = ½m₁v₁² = ½m₂v₂²

= m₁/m₂= (v₂/v₁)²

So, m₁ > m₂

m₁/m₂ >1 

So, that v₂²/v₁²>1

v22>v12

v₂>v₁

So the result shows that the velocity of lighter body v₂ is greater than heavier body v₁.

Therefore, option 3 is correct.

Additional InformationPotential Energy

• The ability of an object to do work due to the position of an object is called potential energy. 
• PE = mgh (where m = mass of an object, g = gravitational acceleration and h = heigh)
• Water collected by building a dam has potential energy.

Relation between Kinetic Energy and Momentum

• K.E = P²/2m

Where, P = Momentum (P = mv)

Important Points Momentum

• The product of the velocity and mass of an object is called momentum.
• The SI unit of momentum is kg-m/s.
• Dimension = MLT^-1

The correct answer is option 2) i.e. Two times more

CONCEPT:

• ​Kinetic energy is the energy possessed by a moving object. Kinetic energy is expressed as

​\(KE = \frac{1}{2} mv^2\)

Where m is the mass of the object and v is the velocity of the object.

EXPLANATION:

• The amount of work required to stop the truck will be equivalent to the kinetic energy possessed by the truck at that moment.
• Work and energy have the same unit i.e. joule and can be equated.

Let m be the mass of the car and v its velocity.

KE of the car = \(\frac{1}{2}mv^2\)

Let m' and v' be the mass and velocity of the truck respectively.

Given that:

m' = 8m and \(v ' =\frac{1}{2}v\)

KE of the truck = \(\frac{1}{2}m'v'^2 = \frac{1}{2}(8m)(\frac{v}{2})^2 = 2 × \frac{1}{2}mv^2\) = 2 × KE of the car

Therefore, the amount of work done required to stop the truck is two times more.

Momentum: A property of a body in motion that is equal to the product of the body's mass and velocity is called momentum.

P = mv

where P is the momentum of the body, m is the mass of the body, and v is the velocity of the body.

Kinetic energy: The energy in a body due to its motion, is known as kinetic energy.

The kinetic energy in terms of momentum is given by:

\(K=\frac{P^2}{2m}\)

where K is the kinetic energy of the body, P is the momentum of the body and m is the mass of the body.

EXPLANATION:

Given that the Momentum of both masses is the same. 

PA = PB

∴ m_AV_A = m_BV_B

∴ \(\frac{m_A}{m_B} = \frac{V_B}{V_A}\)_______(1)

Hence m_A is greater than m_B so the ratio will be more than 1.

Now taking the ratio of the kinetic energy of A and B.

∴ \(\frac{K.E_A}{K.E_B}\) = \(\frac{0.5\times m_A \times V^2_A}{0.5\times m_B \times V^2_B}\) = \(\frac{V_A}{V_B}\) [∵ from eq (1)]

∴  As we know \(\frac{V_A}{V_B}\) has a value lower than 1 from eq 1 so the Kinetic energy of A must be lower than the Kinetic energy of B. So, B has a higher value of kinetic energy.

CONCEPT:

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

W = K_f - K_i

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

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

CALCULATION:

Given – m = 8 kg, F = 16 N, u = 0 m/s and t = 10 s

According to Newton’s second law of motion,

\(a = \frac{F}{m}\)

\(\Rightarrow a = \frac{{16}}{8} = 2\;m/{s^2}\)

From equation of kinematics,

v = u + at

⇒ v = 0 + 2 × 10 = 20 m/s

Kinetic energy acquired by the block is,

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

\(\Rightarrow KE = \frac{1}{2} \times 8 \times {\left( {20} \right)^2} - \frac{1}{2} \times 8 \times {\left( 0 \right)^2} = 1600\;J\)

So option 2 is correct.

The correct answer is option 3) i.e. 6400 J

CONCEPT:

• Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

​Work done, \(W = Δ KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)

Where m is the mass of the object, v is the final velocity of the object and u is the initial velocity of the object. 

CALCULATION:

Given that:

Mass, m = 10 g = 0.01 kg

Initial velocity, u = 1200 m/s

Final velocity, v = 400 m/s

The loss in energy = Work done

\(⇒ Δ KE = W\)

\(⇒​​​​ W=\frac{1}{2}m(v^2 -u^2)\)

\(⇒ ​​​​W=\frac{1}{2}\times 0.01\times(1200^2 -400^2)\)

⇒ W = 6400 J

Concept:

Kinetic energy is defined as the energy that is produced by an object due to its motion.

• The kinetic energy of a moving body, \(K.E.= \frac12mv^2\)
• Where, m = mass of the body, v = velocity of the body
• \(\frac{dv}{dt} = a\)
• F = ma
• Where, m = mass, a = acceleration, 

Explanation:

Given, K = 4t² 

Then \(\frac 12 mv^2 = 4t^2\)

= v² ∝  t²

= v ∝ t

Differentiating both sides,

\(⇒ \frac{d(\frac 12 mv^2)}{dt} = \frac{d(4t^2)}{dt}\)

\(⇒ \frac 12 m(2v)\frac{dv}{dt} = 8t\)

⇒ m.v.a = 8t 

⇒ F.v = 8t

⇒ v ∝  t

⇒ F ∝ constant

So, force acting on particles constant with time.

CONCEPT:

• Work done: The product of the force applied on an object and displacement made by the object in direction of force is known as work done.
• Unit of Work done is Joule which is also unit of Energy.
• Kinetic Energy: The Energy passed by an object by virtue of its motion is called Kinetic Energy.

The kinetic energy of an object with mass m and velocity v is given as:

\(K = \frac{1}{2} mv^{2}\)

• Work-Energy Theorem: Work done on an object is equal to the change in Kinetic Energy of it. This theorem is named as Work-Energy Theorem. 

W.D = (FInal Kinetic Energy) - (Initial KInetic Energy)

\(W.D = \frac{1}{2} mv^{2} - \frac{1}{2} mu^{2} = \frac{1}{2}m(v^{2} - u^{2})\)

u = initial velocity, v = final Velocity, m = mass of Object.

CALCULATION:

Given Initial Velocity u =  20ms^-1

Final Velcocity v = 10 ms^-1

Mass of Object m = 0.5 kg

By Work Energy Theorem

Work done = \(W.D = \frac{1}{2}0.5kg((10ms^{-1})^{2} - (20ms^{-1})^{2})\)

\(W.D = \frac{1}{2}0.5(100 - 400)J\)

⇒W. D = - 300J/4

⇒W. D = - 75J 

So, the Magnitude of the Work done is 75 Joule.

Hence option 1 is correct.

Additional Information

• The expression of Work done is given as W = F.S Cosθ 

θ is the angle between Force applied and displacement made. 

• If both Force and displacement are in the same direction. 
• In that case,  then θ = 0°,  Cos0°, and Work done = FS. That is, simply the product of mass and displacement. 
• If the direction of work done and displacement is opposite, then work done is negative.

Concept:

Kinetic energy: 

• The energy that an object or a particle has by reason of its motion.
• It is a property of a moving object or particle and depends not only on its motion but also on its mass.
• For example, moving car bullet From a gun flying airplane.

The kinetic energy of an object of mass M is given by 

\(K.E = \frac{1}{2}MV^2\)

Potential energy:

• The energy that an object has because of its position relative to other objects.
• It is the property of the object in rest and it also has the potential to convert in another form.
• For example a raised weight, water that is behind a dam, and energy stored in spring.

The potential energy of spring is given by 

\(P.E = \frac{1}{2}Kx^2\)

Where 

K = Spring constant in N/m

x = displacement equilibrium position

Calculation:

Given that

Mass of body m = 2 kg

Velocity of body V = 10 m/s

Spring force constant K = 3200 N/m

When the body strike-through spring, its kinetic energy will be converted into the potential energy of spring. Therefore Loss in KE of body = Gain in potential energy of spring.

⇒ \(\frac{1}{2}mV^2 = \frac{1}{2}Kx^2\)

⇒ x = 0.25 m

Hence, compression produces in spring is 25 cm.

CONCEPT: 

Friction

• The resistance offered by the surfaces that are in contact with each other when they move over each other is called friction.
• Factors affecting friction:
  1. ​​Types of surfaces in contact.
  2. The normal force between the two surfaces.

​\(\Rightarrow F=μ R\)     

where F = friction force, μ = coefficient of friction and R = normal reaction

Work-Energy Theorem: 

• The work-energy theorem states that the net work done by the forces on an object is equal to the change in its kinetic energy.

​⇒ W = ΔKE   

Where W = work done and ΔKE = change in kinetic energy

• Kinetic Energy: The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,

​\(⇒ KE=\frac{1}{2}m× v^{2}\)   

where m = mass and v = velocity

​CALCULATION:

Given v = 20 m/s and μ = 0.4

Initial KE,

\(\Rightarrow KE_{1}=\frac{1}{2}mv^{2}\)     

Final KE, (Final velocity = 0 m/s)

\(\Rightarrow KE_{2}=\frac{1}{2}m\times0^{2}\)

\(\Rightarrow KE_{2}=0\)    

Change in KE,

\(\Rightarrow Δ KE=KE_{2}-KE_{1}\)

\(\Rightarrow Δ KE=\frac{1}{2}mv^{2}\)     

Work is done by the frictional force,

\(\Rightarrow W=F.s\)    

Where, s = displacement

By inserting equation F = μR in the equation, we get,

\(\Rightarrow W=μ Rs\)

\(\Rightarrow R=mg\)

So,

\(\Rightarrow W=μ mgs\)     

As we know, W = ΔKE then,  

\(\Rightarrow μ mgs=\frac{1}{2}mv^{2}\)

\(\Rightarrow 0.4 \times10\times s=\frac{1}{2}20^{2}\)

\(\Rightarrow s=50m\)

• Hence, option 1 is correct.

The correct answer is option 1) i.e. The final kinetic energy increases by the amount of the work

CONCEPT:

• Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

​Work done, \(W = Δ KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)

Where m is the mass of the object, v is the final velocity of the object and u is the initial velocity of the object. 

EXPLANATION:

We know that, \(W = Δ KE =KE_{final} - KE_{initial}\)

• So, if the work done is positive, the final kinetic energy increases by the amount of the work.