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

Calculation:
(I) Given: 0 = 100v + 50(v + 3) (COLM) v = -1 m/s So, work done by man is:

W = (1/2)(100)(1)² + (1/2)(50)(2)² = 150 J   

(II) W_W + W_mg = Δ K.E. (WET) W_N = (1/2)(2)(10)² - (1/2)(2)(5)² = -100 J

(III) Unstable equilibrium points are x = 10 m, 30 m. Stable equilibrium point is x = 20 m. So, U_i = 0, and U_f = -(1/100)(10)⁴ = -100 J. Work done by conservative force is: W_{conservative force} = U_i - U_f = 0 - (-100) = 100 J

(IV) W_ext = U_f - U_i = mg( (5R/8) - (3R/8) ) = mgR/4 = 100 J

The correct answer is option 4) i.e. \(ln(\frac{b}{a})\)

CONCEPT:

• Work is said to be done by a force when the force acting on it causes the object to displace.

Mathematically it is given by;

W = F.x = Fxcosθ 

Where F is the force acting on the object and x is the displacement caused.

• The force acting on an object and displacement is represented graphically as shown. 
  • Consider a varying force acting on the object. If we divide the region under the curve into infinitesimally small regions, the force would appear constant for that region which has caused a displacement of Δx. 
  • In such a case, the area of that small region = Force × displacement (Δx) = work done.

Therefore, work done by a variable force is given by:

​\(W = ∫ _{x_1} ^{x_2} F(x)dx\)

EXPLANATION:

Let x be the distance covered. Given that \(F \propto \frac{1}{x}\)

Work done, \(W = ∫ _{a} ^{b} F(x)dx\)

\(\Rightarrow W \propto ∫ _{a} ^{b} \frac{1}{x}dx\)

\(\Rightarrow W \propto [ln(x)]_a ^b\)

\(\Rightarrow W \propto ln(b)-ln(a)\)

\(\Rightarrow W \propto ln(\frac{b}{a})\)

Calculation:

\(\vec{\tau}=\vec{\mathrm{r}} \times \vec{\mathrm{F}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array}\right| \)

\(\vec{\tau}=\hat{\mathrm{k}}(-1-1)=-2 \hat{\mathrm{k}}\)

\(|\vec{\tau}|=2 \mathrm{Nm}\)

Calculation:

W_all = Δk

W_e = k_f – k_i

\(\mathrm{qE} \frac{\ell}{2}=\frac{1}{2} \mathrm{mv}^{2}-0\)

\(\mathrm{v}=\sqrt{\frac{\mathrm{qE} \ell}{\mathrm{~m}}}\)

Calculation:

Work done

\(\int_{0}^{4} x^{2}(10-x) d x+\int_{0}^{2} y^{2} d y\)

= \(\left[\frac{10 x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{4}+\left[\frac{y^{3}}{3}\right]_{0}^{2}=\frac{640}{3}-64+\frac{8}{3}=152\)

The correct answer is Zero.

Key Points

• When an object rotates in a circle the work done is zero because, in a circular motion, the displacement is always perpendicular to the applied force.
• Also, when a body moves along a circle, the centripetal force acts along the radius of the circle and is perpendicular to the motion of the body.

Important Points

• Work done:
  • work done is defined as the multiplication of the magnitude of the displacement d and the component of the force in the direction of the displacement.
  • Work done through force (F) is equal to a change in kinetic energy.

CONCEPT:

• Work done: It is the dot product of Force and Displacement.

\(W = \overrightarrow{F}.\overrightarrow{d}=Fd~cosθ\)

where W is the work done, F is the force, d is the displacement, and θ is the angle between F and d.

CALCULATION:

Given that W = 1J; F = 10N; d = 10cm = 0.1m; 

Work done is given by 

\(W = Fd~cosθ\)

\(1 = 10\times 0.1~cosθ\)

1 = cosθ 

cos 0° = cosθ

θ = 0° 

• So the angle between force and displacement is 0°
• Hence the correct answer is option 4.

CONCEPT:​

• Work done: It is the dot product of Force and Displacement.

⇒ Work Done (W) = Force (F) × Displacement (S) × cos θ 

where θ is the angle between force and displacement.​

• The gravitational force or weight of a body always acts in the downward direction.

EXPLANATION:

• When a car is moving in a horizontal path, at any particular time,
  • The direction of gravity force is downwards and
  • The direction displacement will be in the direction of motion that is the horizontal direction (perpendicular to the downward direction).

• So the angle between gravity force and displacement will be θ = 90.

⇒ Work Done (W) = Gravity Force (F) × Displacement (S) × cosθ 

⇒ W = F × S × cos90°

⇒ W = F × S × 0 = 0

• So the correct answer is option 1.

Option 2 is correct, i.e. W = F.S cosθ.

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

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

EXPLANATION:

Work done (W) = F.s cosθ 

So option 2 is correct.

CONCEPT:

• Work: Work is said to be done by a force on an object if the force applied causes a displacement in the object.
• The work done by the force is equal to the product of force and the displacement in the direction of the force.
• Work is a scalar quantity.
• Its SI unit is Joule(J).

\(\Rightarrow W=Fx\times cosθ\)    

In vector form,

\(\Rightarrow W=\overrightarrow{F}.\overrightarrow{x}\)     

Where W = work done, F = force, x = displacement and θ = angle between F and x

CALCULATION:

Given \(\vec{F}=(5\hat{i}+3\hat{j})\) and \(\overrightarrow{x}=(2\hat{i}-\hat{j})\)

The work done by the force is:

\(\Rightarrow W=\overrightarrow{F}.\overrightarrow{x}\)

\(\Rightarrow W=(5\hat{i}+3\hat{j}).(2\hat{i}-\hat{j})\)

\(\Rightarrow W=10-3\)

\(\Rightarrow W=7J\)

Hence, option 3 is correct.

CONCEPT:

• Centripetal Force: The force that makes a body to move in a circular motion is known as centripetal force.
  • The direction of this force is always towards the center.

The centripetal force acting on a body of mass 'm' revolving with radius 'r' is:

\(F = {mv^2\over r}\)

• Work done: It is the dot product of Force and Displacement.

Work Done (W) = Force (F) × Displacement (S) × cosθ 

where θ is the angle between force and displacement.​

EXPLANATION:

• When a body is moving in a circular path, at any particular time,
  • The direction centripetal force is towards centre and
  • The direction displacement will be in the direction of motion that is perpendicular to the centripetal direction.

• So the angle between centripetal force and displacement will be θ = 90.

Work Done (W) = Centripetal Force (F) × Displacement (S) × cosθ 

W = F × S × cos90°

W = 0

• So the correct answer is option 3.

CONCEPT:

• Work (W) is said to be done by a force when the force acting on it causes the object to displace.
• Mathematically it is given by:

 \(⇒ W = \vec F .\vec d = Fd\ cos θ \;\;\;\;\; \ldots \left( 1 \right)\) 

Where F = The magnitude of the force vector, d = The magnitude of the displacement vector

EXPLANATION:

Given: The pull force F is applied in the antiparallel direction to the displacement of the mass on the plane.

• Referring to the diagram,

• So, the displacement and pulling force vectors will be in the antiparallel direction to each other as the angle between them will be 180°.

∵ θ  = 180°.⇒ cos180° = -1 

• So, using equation 1,
• The work done in pulling the mass M  will be negative as the magnitude of the force and displacement vector can not be negative.

So, the correct answer will be option 4.

Concept:

Elastic potential energy: It is the potential energy created when an object undergoes an elastic deformation due to a force applied to it.

• This energy is stored in the object as long as the force is in action and the object goes back to its original shape when the force is removed.
• A spring undergoes a similar situation when a force acting on it causes a displacement due to elastic deformation.

The potential energy (U) required to stretch a string by x distance is given by

\(\Rightarrow U = \frac{1}{2}kx^2\)

where k is the spring constant.

Calculation:

Given:

Let the potential energies of the spring be U₁ and U₂ when they are stretched be x₁ and x₂ respectively.

Given that: U₁ = U

When x₁ = 2 cm:

\(\Rightarrow U_1 = \frac{1}{2}kx_1^2 = \frac{1}{2}k(2)^2\)    ------ (1)

When x₂ = 10 cm:

\(\Rightarrow U_2 = \frac{1}{2}kx_2^2 = \frac{1}{2}k(10)^2\)    ------- (2)

On dividing equations (1) and (2), we get

\(\Rightarrow \frac{U_1}{U_2} = \frac{\frac{1}{2}k(2)^2}{\frac{1}{2}k(10)^2} = \frac{2^2}{10^2} = \frac{1}{25}\;\)

⇒ U₂ = 25U₁ = 25U