work-done-by-a-constant-force MCQs for UPSC Prelims

26 practice questions on work-done-by-a-constant-force from the Work, Energy, and Simple Machines section of the UPSC Prelims syllabus. 26 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

8 Easy 13 Medium 5 Hard

Sample questions

Q1
medium
Assertion (A): A man who pushes hard on a stationary wall does a large amount of work on the wall. Reason (R): The work done by a force is zero when the object on which it acts has no displacement.
  1. A Both A and R are true, and R is the correct explanation of A
  2. B Both A and R are true, but R is not the correct explanation of A
  3. C A is true, but R is false
  4. D A is false, but R is true
Show answer and explanation

Correct answer: D - A is false, but R is true

(A) is false: however hard he pushes, the wall does not move, so the work done on it is zero. (R) is a correct statement of when work vanishes, and it is exactly the rule that shows (A) to be wrong. So a true reason sits beside a false assertion.
Q2
hard
A lift carrying passengers is moving downwards at a steady speed. The cable pulls upwards on the lift while gravity pulls it downwards. Which statement about the work done on the lift during this descent is correct?
  1. A Both the cable's pull and gravity do positive work
  2. B Both do zero work, because the lift moves at a steady speed
  3. C The cable's pull does positive work and gravity does negative work
  4. D The cable's pull does negative work and gravity does positive work
Show answer and explanation

Correct answer: D - The cable's pull does negative work and gravity does positive work

The lift moves down, so the downward pull of gravity acts along the displacement and does positive work, while the upward pull of the cable opposes the displacement and does negative work. Swapping them would fit a lift going up. The two forces point in opposite directions, so their work cannot share one sign. A steady speed means the two amounts cancel out, not that each is zero.
Q3
medium
A parcel weighs $2\ \text{N}$. It is raised slowly and steadily straight upwards, so the lifting force is also $2\ \text{N}$. If $6\ \text{J}$ of work is done in lifting it, through what height has it been raised?
  1. A $0.33\ \text{m}$
  2. B $3\ \text{m}$
  3. C $8\ \text{m}$
  4. D $12\ \text{m}$
Show answer and explanation

Correct answer: B - $3\ \text{m}$

Since work is force times height, the height is $6\ \text{J} \div 2\ \text{N} = 3\ \text{m}$. Multiplying instead of dividing gives $12\ \text{m}$. Dividing the force by the work gives about $0.33\ \text{m}$. Adding the two numbers gives $8\ \text{m}$, which is not how work, force and height are related.
Q4
easy
A constant force acts on a box and the box moves. To find the work done by that force, which two quantities must be multiplied together?
  1. A The force, and the speed of the box
  2. B The force, and the time for which it acts
  3. C The mass of the box, and the displacement of the box
  4. D The force, and the displacement measured along the direction of the force
Show answer and explanation

Correct answer: D - The force, and the displacement measured along the direction of the force

Work done by a constant force is that force multiplied by the displacement along its own direction. Time never appears in the definition of work, so force times time is a different quantity. The mass decides how the box responds to a force, not how much work a given force does. Speed tells how fast the box moves, not how far it has been displaced.
Q5
hard
A packing case is hauled $8\ \text{m}$ along a level warehouse floor. A winch cable pulls it forwards along the floor with a constant force of $70\ \text{N}$, while friction opposes the motion with a constant force of $25\ \text{N}$. How much work is done by the net force on the case?
  1. A $760\ \text{J}$
  2. B $200\ \text{J}$
  3. C $360\ \text{J}$
  4. D $560\ \text{J}$
Show answer and explanation

Correct answer: C - $360\ \text{J}$

The net force is $70\ \text{N} - 25\ \text{N} = 45\ \text{N}$ forwards, so the work it does is $45\ \text{N} \times 8\ \text{m} = 360\ \text{J}$. Using the rope's force alone gives $560\ \text{J}$, and using friction's force alone gives $200\ \text{J}$. Adding the two forces instead of subtracting gives $760\ \text{J}$, but friction acts against the pull.
Q6
easy
A constant force of $25\ \text{N}$ acts on a trolley, and the trolley moves $4\ \text{m}$ in the direction of the force. How much work does the force do?
  1. A $29\ \text{J}$
  2. B $50\ \text{J}$
  3. C $100\ \text{J}$
  4. D $6.25\ \text{J}$
Show answer and explanation

Correct answer: C - $100\ \text{J}$

Work is force times displacement along the force: $25\ \text{N} \times 4\ \text{m} = 100\ \text{J}$. Adding the numbers gives $29\ \text{J}$ and dividing them gives $6.25\ \text{J}$, and neither is the rule for work. Halving the product gives $50\ \text{J}$, which has no basis in the given data.
Q7
medium
A book of weight $8\ \text{N}$ is lifted straight upwards through $1.5\ \text{m}$ at a steady speed. What are the signs of the work done by the lifting force and of the work done by gravity on the book?
  1. A Both forces do positive work
  2. B The lifting force does negative work and gravity does positive work
  3. C Both do zero work, because the speed is steady
  4. D The lifting force does positive work and gravity does negative work
Show answer and explanation

Correct answer: D - The lifting force does positive work and gravity does negative work

The lifting force points up and the book moves up, so that work is positive; gravity points down while the book moves up, so its work is negative. The two cannot both be positive, since they point in opposite directions along the same line. Swapping the signs would describe a book being lowered. A steady speed means the two amounts of work cancel, not that each one is zero.
Q8
medium
A student wants to write the joule in terms of the unit of force and the unit of length. Which of these is the same as one joule?
  1. A $1\ \text{N}\cdot\text{m}$
  2. B $1\ \text{N}\cdot\text{s}$
  3. C $1\ \text{kg}\cdot\text{m}$
  4. D $1\ \text{N/m}$
Show answer and explanation

Correct answer: A - $1\ \text{N}\cdot\text{m}$

Work is force times displacement, so its unit is the newton multiplied by the metre: $1\ \text{J} = 1\ \text{N}\cdot\text{m}$. Dividing the newton by the metre reverses the rule and does not give work. Multiplying by the second brings in time, which does not appear in work at all. The kilogram metre mixes mass with length and is not a unit of work.

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