Work energy theorem MCQ Quiz - Objective Question with Answer for Work energy theorem - Download Free PDF

Calculation:
Applying the work-energy theorem (Total work done = Change in mechanical energy):

Calculation:

Work done by all forces equals the change in kinetic energy (KE):

Final velocity vector after 4 seconds:
v = v₀ + a × t = (2i + 2j) × 4 = (8i + 8j)

Magnitude of final velocity:
|v| = √(8² + 8²) = 8√2

Kinetic Energy gained:
KE = (1 / 2) × 2 × (8√2)² = 128 J

Hence, work done by all forces = 128 J

The correct statement regarding the forces acting on the box is: A pair of balanced forces act on it
Explanation:

• A box of mass 500 gm (or 0.5 kg) lying on a horizontal table experiences two main forces: the gravitational force acting downward and the normal force acting upward from the table.
• The gravitational force (weight) acting on the box can be calculated using the equation F = m × g, where m is the mass and g is the acceleration due to gravity.
• Therefore, the gravitational force acting downward is:

F = m × g = 0.5 kg ×  10 m/s² = 5 N

• This means there is a downward force of 5 N due to gravity.
• To prevent the box from moving through the table, the table exerts an upward force known as the normal force. For the box to remain in equilibrium (not accelerating), this normal force must be equal in magnitude and opposite in direction to the gravitational force acting on the box.
• Thus, there's also an upward normal force of 5 N acting on the box.
• Since these two forces (the gravitational force downward and the normal force upward) are equal in magnitude and opposite in direction, they balance each other out.
• This means that there are indeed a pair of balanced forces acting on the box, which keeps it in a state of static equilibrium (not moving) on the table.

Concept Used:

The work done in moving a charge in an electrostatic field depends on the potential difference between the initial and final points.

Since points A, B, and C are on a semicircle and charge q is at the center, they all have the same potential.

Work done in moving a charge between points at the same potential is equal.

Calculation:

Given:

Points A, B, and C are equidistant from charge q at the center.

⇒ V_A = V_B = V_C

⇒ Work done in moving a charge from P to A, B, or C is the same.

⇒ W_A = W_B = W_C

Since moving from P to A, B, or C requires work, the work done is not zero.

∴ The correct answer is W_A = W_B = W_C ≠ 0.

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-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

EXPLANATION:

• According to the work-energy theorem, the work done will be equal to the change in kinetic energy.
• Hence option 3 is correct.

The correct answer is option 4) i.e. for all type of forces

CONCEPT:

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

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

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:

• Energy and work are scalar quantities.
  • So, their values depend on the sum of individual work done by the forces. If these forces cause a displacement, it will be accounted for in work done.
• The work-energy theorem can be applied in the cases of conservative forces, non-conservative forces, internal forces, external forces, and so on.
• Thus, the work-energy theorem is valid for all types of forces.

Concept:

Work

• 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

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

Calculation:

Given, 

Change in Kinetic energy = ΔKE = 3 kJ = 3000 J

Force is along the line of motion F = 1000 N

The angle between the force and displacement θ = 0° 

By work energy theorem

W = Fs cos θ = 3000 J

1000 N × s cos 0° = 3000 J

1000 N × s = 3000 J

s = 3 m

So displacement is 3 m. 

CONCEPT:

• Kinetic Energy: When a body is in linear motion, its energy due to motion will be kinetic energy. Mathematically

K = 1/2 × m × v2

where K is the kinetic energy of the body, and m is the mass of the body, and v is its velocity.

• Work-Energy Theorem: The net work done on an object is equal to a change in its Kinetic energy.

W = ΔK

W = Kf -Ki 

where W is work done, K_f is final kinetic energy and K_i is initial kinetic energy.

CALCULATION:

Given that initial velocity = v

final velocity = 2v

Let the mass of the truck is m

so initial kinetic energy K_i = 1/2 × m × v2

final kinetic energy K_f = 1/2 × m × (2v)2 = 2 × m × v²

Net work done W = ΔK = Kf -Ki =  2 × m × v2 -  1/2 × m × v²

W = 3/2 × m × v2

W = 3 K_i

So Net work done is three times as the initial kinetic energy of the mass or work done in accelerating it from rest to v. Hence the correct answer is option 1.

The correct answer is option 2) i.e. the net work done by the force on the object

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, \(W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)

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:

• ​From the work-energy theorem, the net work done by the forces on an object equals the change in its kinetic energy.
• Hence, the correct answer is option 2).

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 = Kf - Ki

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

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

CALCULATION:

Given - mass (m) = 120 kg, initial velocity (u) = 0 km/hr = 0 m/s and final velocity (v) = 18 km/hr = 5 m/s

• The work done by the boy is

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

\(⇒ W = \frac{1}{2}\times 120\times ({25} - {0} )=1500\, J\)

Concept:

• Kinetic Energy: When a body is in linear motion, its energy due to motion will be kinetic energy. Mathematically

⇒ K = 1/2 × m × v²

Where K is the kinetic energy of the body, m is the mass of the body, and v is its velocity.

• Work-Energy Theorem: The net work done on an object is equal to the change in its Kinetic energy.

⇒ W = ΔK

⇒ W = K_f  - K_i 

Where W is work done, K_f is final kinetic energy and K_i is initial kinetic energy.

Explanation:

• Since the ball rebounds with the same velocity. Therefore the change in kinetic energy is zero. Hence, the work done by the ball on the wall is zero

Concept:

• The work done by a force is defined as: The product of the component of the applied force on the block in the direction of the displacement and the magnitude of this displacement is called work done.
• The work-energy theorem is also called the principle of work and kinetic energy.
• The work-energy theorem states that the total work done by all the forces is equal to the change in kinetic energy of the particle.
• Total work done = change in kinetic energy

⇒ Total work done (W) = KE_f – KE_i

\(⇒ W = \frac{1}{2}m\;{v^2} - \frac{1}{2}m\;{u^2}\)

Where KE_f is final kinetic energy, KE_i is initial kinetic energy, u is initial velocity and V is the final velocity

• For a variable force, the work-energy theorem is given by:

\(⇒ Work\;done\;\left( W \right) = \mathop \smallint \nolimits_{{x_1}}^{{x_2}} F.\;dx = \frac{1}{2}m\;{V^2} - \frac{1}{2}m\;{v^2}\)

Explanation:

• From the above-given explanation, we can see that the relation between kinetic energy and work is given by the work-energy theorem
• And according to the work-energy theorem Work done, W = ΔK.E
• From this, we can see that if work done on an object increases then the Kinetic energy associated with it will also increase. Hence option 2 is correct