Class 9 Work, Energy, and Simple Machines Mock Test (CBSE)

Topics in Work, Energy, and Simple Machines

• Work done by a constant force
• Zero, positive and negative work
• Work-energy theorem
• Forms of energy
• Kinetic energy
• Gravitational potential energy
• Conservation of mechanical energy
• Power
• Simple machines: pulley, inclined plane, lever

Work, Energy, and Simple Machines practice questions with answers

1. Q1Multiple choice·1 mark
   A particle of mass $1\text{ kg}$ experiences a force $F = -kx$ where $k = 100\text{ N/m}$. If it is released from rest at $x = 0.2\text{ m}$, its maximum speed during the motion is:

   1. (a)$1.0\text{ m/s}$
   2. (b)$2.0\text{ m/s}$
   3. (c)$0.5\text{ m/s}$
   4. (d)$1.41\text{ m/s}$
   Show answer and step marks

   1. Correct option: (b) $2.0\text{ m/s}$. Apply conservation of mechanical energy: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2 \implies v = x\sqrt{k/m} = 0.2 \times \sqrt{100/1} = 2.0\text{ m/s}$.+1

   Answer: (b)
2. Q2Multiple choice·1 mark
   An engine lifts a mass of $200\text{ kg}$ to a height of $10\text{ m}$ in $5\text{ s}$. What is the power developed by the engine? (Take $g = 10\text{ m/s}^2$)

   1. (a)$400\text{ W}$
   2. (b)$2000\text{ W}$
   3. (c)$10000\text{ W}$
   4. (d)$4000\text{ W}$
   Show answer and step marks

   1. Correct option: (d) $4000\text{ W}$+1

   Answer: (d)
3. Q3Multiple choice·1 mark
   Which of the following is an example of a class-II lever?

   1. (a)Wheelbarrow
   2. (b)Tongs
   3. (c)Scissors
   4. (d)Human forearm
   Show answer and step marks

   1. Correct option: (a) Wheelbarrow+1

   Answer: (a)
4. Q4Short answer·2 marks
   A constant horizontal force of $35\text{ N}$ is applied to push a crate across a workshop floor, producing a displacement of $4\text{ m}$ in the direction of the applied force. (a) Calculate the work done by the force. (b) State the SI unit of work and whether work is a scalar or vector quantity.
   Show answer and step marks

   1. Using the formula $W = F \times s$ and substituting values: $W = 35\text{ N} \times 4\text{ m} = 140\text{ J}$.+1
   2. The SI unit of work is the joule ($\text{J}$).+0.5
   3. Work is a scalar quantity.+0.5

   Answer: Work done = $140\text{ J}$, SI unit is Joule, scalar quantity.
5. Q5Short answer·3 marks
   A ball of mass $0.2\text{ kg}$ is thrown vertically upwards with an initial velocity of $20\text{ m/s}$. (Take $g = 10\text{ m/s}^2$) (i) Calculate the work done by the gravitational force during the upward journey until the ball reaches its maximum height. (ii) State whether this work is positive or negative, and give a reason. (iii) What is the total work done by the gravitational force during the complete round trip (upward and downward journey back to the point of projection)?
   Show answer and step marks

   1. Maximum height $h = \frac{u^2}{2g} = \frac{20^2}{2 \times 10} = 20\text{ m}$. Work done by gravity $W = -mgh = -0.2 \times 10 \times 20 = -40\text{ J}$.+1
   2. The work is negative because the gravitational force acts downwards while the displacement of the ball is upwards (angle between force and displacement is $180^\circ$).+1
   3. Total work done for the complete round trip is zero because the net displacement is zero ($W_{\text{total}} = -40\text{ J} + 40\text{ J} = 0\text{ J}$).+1

   Answer: (i) $-40\text{ J}$. (ii) Negative, because gravity acts downward while the ball's displacement is upward (angle $180^\circ$). (iii) $0\text{ J}$, because the ball returns to the point of projection, so the net displacement is zero.
6. Q6Short answer·3 marks
   A body of mass $2\text{ kg}$ is initially moving in a straight line with a velocity of $4\text{ m/s}$. A constant net force acting along its direction of motion increases its velocity to $10\text{ m/s}$ over a displacement of $7\text{ m}$. Using the work-energy theorem, find the magnitude of the net force acting on the body.
   Show answer and step marks

   1. According to the work-energy theorem, net work done equals change in kinetic energy: $W = \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.+1
   2. Substitute the given values: $W = \frac{1}{2} \times 2 \times (10^2 - 4^2) = 1 \times (100 - 16) = 84\text{ J}$.+1
   3. Equate work to force times displacement: $W = F \times s \implies 84 = F \times 7$, which gives $F = 12\text{ N}$.+1

   Answer: $12\text{ N}$

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