Work MCQ Quiz - Objective Question with Answer for Work - 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

Concept:

The work done in stretching a wire is given by the formula: W = (1/2) × F × Δx, where F is the force applied and Δx is the extension. The force required to stretch a wire is related to its Young's Modulus (Y) and the strain produced:

F = Y × A × Δx / L, where A is the cross-sectional area, L is the length, and Δx is the extension.

Calculation:

Given, for the first wire:

Work done, W₁ = 2 J

The second wire has double the radius (2r) and half the length (L/2). The cross-sectional area is proportional to the square of the radius, so the area of the second wire becomes four times that of the first wire (A₂ = 4A). The force required to stretch the wire is inversely proportional to the length, so the force for the second wire is half that of the first wire (F₂ = ½F₁). Since the work is proportional to the force and the square of the radius, the work done for the second wire will be:

W₂ = 2 J × (4 × 2) = 16 J

∴ The work necessary to stretch the second wire is 16 J. Option 1) is correct.

The correct answer is +1 J.

Key Points

• Work Done is defined as the product of the force applied and the displacement in the direction of the force.
• The formula for work done is W = F × d, where W is the work done, F is the force, and d is the displacement.
• In this case, the force (F) is 1 N and the displacement (d) is 1 m.
• Therefore, the work done (W) is 1 N × 1 m = 1 J.
• The positive sign indicates that the force and the displacement are in the same direction.

 Additional Information

• Negative Work Done
  • Negative work occurs when the force and the displacement are in opposite directions.
  • In such cases, the work done is calculated as -F × d.
  • This situation typically occurs in scenarios involving friction or opposing forces.
• Zero Work Done
  • Zero work is done when either the force is zero, or the displacement is zero, or the force is perpendicular to the direction of displacement.
  • In mathematical terms, this can be expressed as W = 0.

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

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

W = Fs cos θ

Where θ = angle between force direction of motion

• Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
• SI unit of work is Joule (J).

CALCULATION:

Given - Force (F) = 20 N, distance (s) = 13 m and θ = 30° 

• The mathematical expression of work done is given by 

⇒ W = Fs cos θ

⇒ W = 20 × 13 × cos 30°                     

⇒ W = 225 Joules

The correct answer is Only I.

Key Points

• The work done by the spring force in a cyclic process is zero.
• This is because in a cyclic process, the system returns to its initial state, meaning the net change in potential energy of the spring is zero.
• When a spring is compressed or stretched, it stores potential energy given by 12kx2" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">12kx212kx2" id="MathJax-Element-1232-Frame" role="presentation" style="position: relative;" tabindex="0">12kx212kx2 12kx2 $\frac{1}{2}kx^2$ , where k" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">kk" id="MathJax-Element-1233-Frame" role="presentation" style="position: relative;" tabindex="0">kk k $k$ is the spring constant and x" id="MathJax-Element-31-Frame" role="presentation" style="position: relative;" tabindex="0">xx" id="MathJax-Element-1234-Frame" role="presentation" style="position: relative;" tabindex="0">xx x $x$ is the displacement from the equilibrium position.
• During a complete cycle, the spring goes through equal amounts of compression and extension, leading to equal and opposite work done by the spring force.
• As a result, the total work done by the spring force over one complete cycle is zero, confirming that the correct answer is only I.

 Additional Information

• Positive Work (Option II)
  • Positive work occurs when the force applied and the displacement are in the same direction. However, in a cyclic process, this is not the case over the entire cycle.
• Negative Work (Option III)
  • Negative work occurs when the force applied and the displacement are in opposite directions. Similar to positive work, this is not the case over the entire cycle.
• Combination of I, II, and III (Option IV)
  • While different segments of the cycle may involve positive or negative work, the net work done over a complete cycle is zero, ruling out the possibility of all three being correct simultaneously.

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:

• 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 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:

• 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

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).

W = F.x cos θ     

In vector form, work is dot product of force and displacement. 

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

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

Explanation:

A mass is moved from one place to another on a horizontal road. 

The force of gravity is acting in a downward direction. 

The angle between displacement and force θ = 90 ° 

W = F.x cos θ 

W = F. x. cos 90 ° = F . x (0) = 0 

as cos 90° = 0

Therefore, the work done is 0 Joule. 

Important Points

Whenever the angle between force and displacement is 90°, the work done is zero. Circular motion is another example. 

Concept:

When a force is applied on an object which causes displacement along the direction of force then the work is said to be done by that force.

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.

i.e., \(W = \vec F.d\vec x = Fdx\cos \theta \)

Here both force and displacement will be along the same direction, since force is cause this displacement thus Work done is a scalar quantity.

And SI unit of Work is Joule whereas CGS unit is erg

• Whereas, 1 joule = 10⁷ erg

Explanation:

From the above explanation we can see that both Joule and Erg are units of Works.

Thus, the conversion from joule to erg can be given as

1 joule = 10⁷ erg

Which means option 2 is correct among all

Note:

Both energy and work have the same unit and dimension, i.e., Joule is S.I unit of both Work and energy 

The correct answer is option 2) i.e. The weight of the luggage is perpendicular to the distance moved

CONCEPT:

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

Mathematically it is denoted by:

W = F.x

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

• Work depends on two factors: Force and displacement.
  • Positive work: If the force applied and displacement are in the same direction, work done is said to be positive.
  • Negative work: If the displacement is in a direction opposite to the direction of applied force, work done is said to be negative.
  • Zero work: If the direction of displacement and force are mutually perpendicular, work done is said to be zero.

EXPLANATION:

• In the given case, the force acting on the luggage is its weight i.e. acting downwards due to gravity.
• As the porter moves from one end of the platform to the other end, the distance moved is perpendicular to the direction of the force.

Thus, the work done on the luggage will be zero as the weight of the luggage is perpendicular to the distance moved.

The correct answer is option 2, i.e 315 W.

⇒ Given, the mass of the body, m = 50 kg. h is the height, t is the time. g is the accelaration due to gravity which is aproximated to 10 m/s2 

⇒ h = 45 × 14 = 630 cm = 6.30 m

⇒ t = 10 s, g = 10 ms-2

⇒ Potential Energy = mgh = 50 × 10 × 6.30 = 3150 J. (where J is the unit of energy/work which is joules)

⇒ Power = PE / time = Energy / time = 3150 / 10 = 315 watt.  (watt is the unit of power which is different from units)

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.

EXPLANATION:

• When the man moves on a horizontal surface, then work done by gravity is zero since gravitational force acts perpendicular to the displacement i.e., θ = 90° 

⇒ W = Fs cos 90° = 0         [∵ cos 90° = 0]

• When the moon is revolving around the earth then the work done by the centripetal force is always zero because centripetal force acts along the radius of curvature and towards its center but the displacement of the body is along circumference (tangent) which is perpendicular to the radius and thus the angle between the centripetal force and displacement is 90°.

⇒ W = Fs cos 90° = 0         [∵ cos 90° = 0]

• When a man travels on a horizontal platform with a load on his head, work done against gravity by the man is zero since gravitational force acts perpendicular to the displacement i.e., θ = 90° 

⇒ W = Fs cos 90° = 0         [∵ cos 90° = 0

• Therefore the work done is zero in all the cases. Therefore option 4 is correct.