This set of Class 11 Physics Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Work”.

1. Work done in a system is ___

a) The measure of energy change

b) The measure of useful work

c) The measure of displacement

d) The measure of force

View Answer

Explanation: Work done in any system is a measure of the energy change in the system with its surroundings. Work done and energy change are in practice the same. When a force is applied on a body and it moves and does some work. The amount of work done denotes energy change of the system with its surroundings.

2. Work is a _______

a) Scalar quantity

b) Vector quantity

c) Foreign quantity

d) Magnitude less quantity

View Answer

Explanation: Work done by a force on a body is a scalar quantity. The work done is obtained by having a dot product of the force vector with the displacement vector. The dot product of any two vectors gives a scalar as the result.

3. A force of 40 N acts on body of mass 5 Kg. What is the amount of work done if the total displacement is 2 m and is in the direction of the force applied?

a) 80 J

b) -80 J

c) 40 J

d) 16 J

View Answer

Explanation: The work done on a body is obtained by calculating the dot product of the force and displacement vector. Since the displacement is in the direction of the applied force, the work done is positive and is equal to 80 J.

4. Work being a scalar quantity sometimes has negative and positive signs. What does the sign suggest?

a) The direction of work

b) Relative directions of force and displacement

c) The direction of force

d) The direction of displacement

View Answer

Explanation: Work is a scalar quantity; hence it cannot have a direction. But work is obtained by calculating the dot product of force and displacement. If both are in the same direction, the dot product is positive. If both are in opposite directions, the dot product is negative. Hence, the sign of work suggests the relative directions of force and displacement.

5. A force of 40 N acts on body of mass 5 Kg which is initially at rest. What is the amount of work done in the first 10 s?

a) 16000 J

b) -16000 J

c) 400 J

d) -400 J

View Answer

Explanation: The work done on a body is obtained by calculating the dot product of the force and displacement vector. From Newton’s second law, we get the acceleration equal to 8 m/s

^{2}. From the second equation of motion the displacement covered can be obtained which is equal to 400 m. On taking the dot product of the force and displacement, we get work done equaling to 16000 J.

6. The work done on a body of mass 5 Kg is 70 J. What is the magnitude of the force applied if the total displacement it covered is 7 m.

a) 10 N

b) -10 N

c) 20 N

d) 70 N

View Answer

Explanation: The amount of work done on a body is obtained by calculating the dot product of the force and the displacement. Due to lack of information, we assume that the force and the displacement are in the same direction. Since the total work done is given, we can find out the applied force by dividing the total work done by the total displacement. By doing so we get force applied = 10 N.

7. A body of mass 40 Kg covers a distance of 50 m in 5s starting from rest under the influence of a force. What is the work done by the force on the body?

a) 8000 J

b) -8000 J

c) 20000 J

d) 800 J

View Answer

Explanation: Since the distance and time taken are given, we can find out the acceleration of the body by the second equation of motion which is 4 m/s

^{2}. From the second law of motion, the force can be found. Force = 160 N. From the force and the displacement, the work can be found out. Work = 8000 J.

8. A block of mass 10 Kg accelerates from rest to a velocity of 20 m/s in 5s under the influence of an external force. What is the work done by the external force?

a) 2000 J

b) -2000 J

c) 20000 J

d) -20000 J

View Answer

Explanation: The acceleration can be found out by using the initial and final velocities and the time taken. Acceleration = 4 m/s

^{2}. Force can be found out by the second law. By the second equation of motion the distance covered in 5 s can be found which is equal to 50 m. Work can be found out by multiplying the force and the distance covered. Work = 2000 J.

9. A force of 10 N creates a displacement of 2 m but the work done on the body throughout its motion is equal to 70 J. What possibly can be the reason or this?

a) There are multiple forces action on the body

b) The body is at rest

c) The force is not uniform

d) The displacement is harmonic

View Answer

Explanation: The possible reason for this is that multiple forces are acting on the body at different instances throughout its motion. The work done on a body is a measure of the energy exchanged between the body and the environment and is usually the total work done. Work done due to the given force is 20 J but this need not be the total work done.

10. Work done by a force on a body is an example of ______

a) Mechanical work

b) Electrical work

c) Magnetic work

d) Thermodynamic work

View Answer

Explanation: The work done by a force on a body is called as mechanical work. In mechanical work, the mechanical energy of the body changes. Mechanical work = W = \(\vec{F}.\vec{s}\), where F is the force and s is the displacement.

**Sanfoundry Global Education & Learning Series – Physics – Class 11**.

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