07. Short and Long Answer Questions: Work, Energy, and Simple Machines

Short Answer Type Questions

Ques 1: Define work done by a constant force. Write its SI unit and state under what two conditions work done on an object is zero.

Ans: The work done by a constant force on an object is equal to the product of the force applied and the displacement of the object in the direction of the force.

$$W=F\times s$$

The SI unit of work is the joule (J). One joule is the work done when a constant force of 1 newton displaces an object by 1 metre in the direction of the force, i.e., $1\text{ J}=1\text{ N}\times 1\text{ m}=1{\text{ kg m}}^2{\text{ s}}^{-2}$.

Work done is zero when:

1. The displacement of the object is zero $(s=0)$, regardless of the force applied – for example, pushing a rigid wall.
2. The force applied is zero $(F=0)$ – for example, an object moving with constant velocity on a frictionless surface has no net force, so no work is done by the net force.

Note: Work has no direction. It can be described using a positive or negative sign.

Ques 2: Distinguish between positive work and negative work done by a force. Give one example of each.

Ans: Positive work is done by a force on an object when the displacement of the object is in the same direction as the applied force. The object gains energy.
Example: When a boy pushes a wheelchair forward and the wheelchair moves in the same direction, the boy does positive work on the wheelchair.

Negative work is done by a force on an object when the displacement of the object is in the direction opposite to that of the force. The object loses energy.
Example: When a goalkeeper stops a ball by applying a force opposite to the direction of the ball’s motion, the goalkeeper does negative work on the ball. The work done $=F\times (-s)=-30\text{ J}$ (negative because force and displacement are in opposite directions).

Work done by friction on a moving object is always negative because friction acts opposite to the direction of motion.

Ques 3: State the work-energy theorem. How does it explain that when positive work is done on an object, its kinetic energy increases?

Ans: The work-energy theorem states that the work done on an object or a system is equal to the change in its energy:

$$\text{Work done on an object}=\text{Change in its energy}$$

This theorem holds even when the forces applied are not constant and for a system of objects.

Explanation: When positive work is done on an object by a net force, the force and displacement are in the same direction. Using $W=F\times s$ and Newton’s second law $F=ma$, with the kinematic equation $v^2=u^2+2as$:

$$W=ma\times s=ma\times \frac{v^2-u^2}{2a}=\frac{1}{2}m(v^2-u^2)$$

Since this work equals the change in energy, the object’s kinetic energy increases from $\frac{1}{2}m{u}^2$ to $\frac{1}{2}m{v}^2$. Thus, positive work increases the velocity and kinetic energy of the object, while negative work decreases them.

Ques 4: Derive the expression for kinetic energy of an object of mass m moving with velocity v. What happens to the kinetic energy if the velocity of the object doubles?

Ans: Consider an object of mass $m$ starting from rest $(u=0)$ and acquiring velocity $v$ under a constant force $F$ over displacement $s$. Using the kinematic equation:

$$v^2=u^2+2as=0+2as\Rightarrow s=\frac{v^2}{2a}$$

Work done by the force: $W=F\times s=ma\times {\displaystyle \frac{v^2}{2a}}={\displaystyle \frac{1}{2}}m{v}^2$

By the work-energy theorem, this work equals the kinetic energy gained:

$$K=\frac{1}{2}m{v}^2$$

Effect of doubling velocity: If velocity changes from $v$ to $2v$:

Original KE $={\displaystyle \frac{1}{2}}m{v}^2$

New KE $={\displaystyle \frac{1}{2}}m(2v{)}^2={\displaystyle \frac{1}{2}}m\times 4v^2=4\times {\displaystyle \frac{1}{2}}m{v}^2$

The kinetic energy becomes 4 times the original value.

Ques 5: What is potential energy? Write the expression for the gravitational potential energy of an object at height h above the ground and derive it using the work-energy theorem.

Ans: The potential energy is the energy stored by an object as a result of its deformation or in a system of objects due to their relative positions. The gravitational potential energy is the potential energy of the Earth-object system due to the height of the object above the ground.

Derivation: To raise an object of mass $m$ slowly from the ground to a height $h$, we apply an upward force equal to the gravitational force $mg$. The work done $W$ by this applied force on the object is:

$$W=\text{force}\times \text{displacement}=mg\times h=mgh$$

By the work-energy theorem, this work appears as the change in potential energy of the object. Since the potential energy at the ground is defined as zero:

$$U=mgh$$

The unit of potential energy is the joule (J), same as kinetic energy and work. The greater the height $h$, the greater the potential energy of the object.

Ques 6: A fielder throws a cricket ball of mass 200 g to a height of 10 m above the ground. Calculate its potential energy at the maximum height. Also find its kinetic energy at the moment it was thrown, assuming no energy loss. Take g=10 m s−2.

Ans: Potential energy at maximum height:

Mass $m=200\text{ g}=0.2\text{ kg}$

Height $h=10\text{ m}$, $g=10{\text{ m s}}^{-2}$

$$U=mgh=0.2\times 10\times 10=20\text{ J}$$

Kinetic energy at the moment of throwing:

At maximum height, the velocity of the ball is zero, so its kinetic energy $=0$. By conservation of mechanical energy (no energy loss), all kinetic energy at the time of throwing is converted to potential energy at maximum height:

Kinetic energy at throw $=$ Potential energy at maximum height

$$\Rightarrow K=20\text{ J}$$

Ques 7: State the law of conservation of mechanical energy. Using a freely falling object, show that the total mechanical energy remains constant at any point during the fall.

Ans: Conservation of Mechanical Energy: The mechanical energy (sum of kinetic energy and potential energy) of an object remains constant when no external forces other than gravity act on it. The total mechanical energy is conserved.

Proof for a freely falling object: Consider an object of mass $m$ dropped from height $h$ (initial velocity $u=0$).

At the top (Point A):

Potential energy $=mgh$, Kinetic energy $=0$

Total mechanical energy $=mgh+0=mgh$

At any intermediate point B (height h′ above ground, after time t):
Using kinematics: $v=gt$ and $h^{\mathrm{\prime }}=h-\frac{1}{2}g{t}^2$

KE $=\frac{1}{2}m{v}^2=\frac{1}{2}m{g}^2{t}^2$

PE $=mg{h}^{\mathrm{\prime }}=mgh-\frac{1}{2}m{g}^2{t}^2$

Total $=mgh-\frac{1}{2}m{g}^2{t}^2+\frac{1}{2}m{g}^2{t}^2=mgh$ ✓

Thus, at every point during the fall, the total mechanical energy remains $mgh$ – it is conserved. The potential energy lost equals the kinetic energy gained.

Ques 8: Define power. Write its formula and SI unit. A weightlifter lifts a 75 kg mass by 2 m in 5 seconds. Calculate the power required. Take g=10 m s−2.

Ans: Power is defined as the rate at which work is done. It tells us how quickly or slowly work is done.

$$P=\frac{W}{t}$$

The SI unit of power is the watt (W), named after James Watt. One watt equals 1 joule of work done per second: $1\text{ W}=1{\text{ J s}}^{-1}$.

Calculation:

Work done $=mgh=75\times 10\times 2=1500\text{ J}$

Time $t=5\text{ s}$

$$P=\frac{W}{t}=\frac{1500\text{ J}}{5\text{ s}}=300\text{ W}$$

Note: To do more work in the same time, or the same work in less time, more power is required.

Ques 9: What is a simple machine? Define mechanical advantage. State the mechanical advantage of a fixed pulley and explain why it is still useful despite having a mechanical advantage of 1.

Ans: A simple machine is a device that makes work easier by changing the magnitude or direction of the force required to perform a task. Although the total work cannot be reduced, a simple machine reduces the effort or makes it more convenient to apply.

Mechanical advantage (MA) is defined as the ratio of the load (force to be overcome) to the effort (force applied by the user):

$$\text{MA}=\frac{\text{Load}}{\text{Effort}}$$

Fixed pulley: In a fixed pulley, the effort and load are equal in magnitude, so the mechanical advantage $=1$. It does not reduce the force required.

Why it is still useful: Even though a fixed pulley has MA = 1, it changes the direction of the effort. It is much easier and more convenient for a person to pull a rope downward (using body weight) than to lift a load directly upward. For example, raising a flag or lifting building materials using a pulley is far more comfortable than a direct vertical lift.

Ques 10: What is an inclined plane? Derive the expression for its mechanical advantage. If a ramp has a length of 5 m and a height of 2 m, calculate its mechanical advantage.

Ans: An inclined plane is a simple machine – a smooth sloping surface – that allows a heavy load to be raised to a height using a smaller effort force applied over a longer distance.

Derivation of Mechanical Advantage:
Let mass of object $=m$, height to be raised $=h$, length of inclined plane $=L$, and effort force $=F^{\mathrm{\prime }}$.

At constant speed, using the work-energy theorem (ignoring friction):

Work done by effort $=F^{\mathrm{\prime }}\times L$

Potential energy gained $=mgh$

Equating: $F^{\mathrm{\prime }}\times L=mgh\Rightarrow {\displaystyle \frac{mg}{F^{\mathrm{\prime }}}}={\displaystyle \frac{L}{h}}$

$$\text{MA}=\frac{\text{Load}}{\text{Effort}}=\frac{mg}{F^{\mathrm{\prime }}}=\frac{L}{h}$$

Calculation:

Length $L=5\text{ m}$, Height $h=2\text{ m}$

$$\text{MA}=\frac{L}{h}=\frac{5}{2}=2.5$$

A longer inclined plane (smaller angle) gives a greater mechanical advantage, meaning a smaller effort is needed – but the object must be moved over a greater distance.

Ques 11: Explain the three parts of a lever. Write the condition for a lever to be balanced and derive the expression for its mechanical advantage.

Ans: A lever is a simple machine consisting of a rigid bar that can rotate about a fixed point. It has three main parts:

1. Fulcrum: The fixed point about which the lever rotates.
2. Load: The force to be overcome (the resistance force $F_2$). The distance of the load from the fulcrum is called the load arm $(d_2)$.
3. Effort: The force applied by the user $(F_1)$. The distance of the effort from the fulcrum is called the effort arm $(d_1)$.

Condition for balance (principle of lever):
Work done at the effort end = Work done at the load end:

$$F_1\times d_1=F_2\times d_2$$$$\text{Effort}\times \text{Effort arm}=\text{Load}\times \text{Load arm}$$

Mechanical Advantage of a Lever:

$$\text{MA}=\frac{\text{Load}}{\text{Effort}}=\frac{F_2}{F_1}=\frac{d_1}{d_2}=\frac{\text{Effort arm}}{\text{Load arm}}$$

By increasing the effort arm, the lever can apply a larger force on the load than the effort. However, a lever reduces the force required, not the total work done.

Ques 12: Name the different forms of energy. Give one example of conversion from one form of energy to another for any three forms you name.

Ans: Energy exists in many forms. The major forms of energy are: mechanical energy (kinetic and potential), thermal energy, light energy, sound energy, electrical energy, chemical energy, and nuclear energy.

Three examples of energy conversion:

1. Chemical energy → Mechanical energy: The chemical energy stored in food is used by our muscles to produce mechanical energy for walking and running. Similarly, chemical energy in fuel powers a car’s engine, producing mechanical energy.
2. Electrical energy → Light and Thermal energy: When electrical energy passes through the filament of a light bulb, it is converted into light energy and thermal energy (heat). An electric water heater converts electrical energy into thermal energy.
3. Mechanical energy → Sound energy: When a bell is struck, mechanical energy (the kinetic energy of the hammer) is converted into sound energy – the vibrations of the bell produce sound waves.

Energy cannot be created or destroyed – it can only be converted from one form to another. The total energy of a closed system remains constant.

Long Answer Type Questions 

Ques 1: What is energy? Define kinetic energy and potential energy. Derive the mathematical expression for kinetic energy from first principles. A car of mass 1000 kg starts from rest and reaches a speed of 72 km h−1 in 10 s. Calculate the kinetic energy of the car and the power of the engine. Take g=10 m s−2.

Ans: Energy is the capacity of an object to do work. An object having energy can exert a force on another object and cause it to move. The SI unit of energy is the joule (J), the same as the unit of work.

Kinetic Energy (K): The energy possessed by an object due to its motion is called kinetic energy. All moving objects possess kinetic energy.

Potential Energy (U): The energy stored by an object as a result of its deformation or due to the relative positions of objects in a system is called potential energy. Examples include a stretched rubber band, a raised object, and a compressed spring.

Derivation of Kinetic Energy:
Consider an object of mass $m$ starting from rest $(u=0)$ under constant force $F$, acquiring velocity $v$ over displacement $s$.

1. From kinematics: $v^2=u^2+2as=0+2as\Rightarrow s={\displaystyle \frac{v^2}{2a}}$
2. Work done by force: $W=F\times s=ma\times {\displaystyle \frac{v^2}{2a}}$
3. Simplifying: $W={\displaystyle \frac{1}{2}}m{v}^2$
4. By the work-energy theorem, this equals the kinetic energy gained:

$$K=\frac{1}{2}m{v}^2$$

Numerical Solution:

Mass $m=1000\text{ kg}$, Initial velocity $u=0$

Final velocity $v=72{\text{ km h}}^{-1}={\displaystyle \frac{72000}{3600}}=20{\text{ m s}}^{-1}$

Kinetic energy: $K={\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{1}{2}}\times 1000\times (20{)}^2=200000\text{ J}=200\text{ kJ}$

Work done by engine $=$ Change in KE $=200000-0=200000\text{ J}$

$$P=\frac{W}{t}=\frac{200000\text{ J}}{10\text{ s}}=20000\text{ W}=20\text{ kW}$$

Ques 2: Explain the conservation of mechanical energy with reference to a simple pendulum. Also explain why in real life the pendulum eventually comes to rest. Using conservation of mechanical energy, find the velocity of a child of mass m at the bottom of a slide of height h (neglect friction).

Ans: Mechanical energy is the sum of kinetic energy and potential energy of an object. The conservation of mechanical energy states that when no external force (other than gravity) acts on an object, its total mechanical energy remains constant throughout its motion.

Conservation in a Pendulum:
Consider a simple pendulum bob swinging between points P, Q, and R, where Q is the bottom-most point and P and R are the extreme positions at height $h$ above Q.

1. At point P (extreme position): The bob momentarily stops. Velocity = 0, so KE = 0. All energy is potential: $\text{Total ME}=mgh+0=mgh$.
2. At point Q (bottom-most position): The bob has maximum velocity $v$. Height = 0, so PE = 0. All energy is kinetic: $\text{Total ME}=0+\frac{1}{2}m{v}^2$. By conservation, $\frac{1}{2}m{v}^2=mgh$.
3. At point R (other extreme): Again velocity = 0, KE = 0, and PE = $mgh$. The bob regains the same height $h$ it started with.

At every point, $\text{KE}+\text{PE}=mgh$ = constant. This demonstrates conservation of mechanical energy.

Why the pendulum eventually stops in real life:
In real life, friction at the support point and air resistance continuously do negative work on the pendulum bob, removing small amounts of mechanical energy at each swing. This lost mechanical energy is converted into thermal energy (heat) and sound. As the mechanical energy decreases, the amplitude reduces and the pendulum eventually comes to rest at the equilibrium point Q.

Velocity of a child at the bottom of a slide:
At the top: $\text{PE}=mgh$, $\text{KE}=0$.
At the bottom: $\text{PE}=0$, $\text{KE}=\frac{1}{2}m{v}^2$.

By conservation of mechanical energy:

$$mgh=\frac{1}{2}m{v}^2\Rightarrow v=\sqrt{2gh}$$

The velocity at the bottom depends only on the height $h$ of the slide, not on the mass of the child or the shape of the slide.

Ques 3: What are simple machines? Name the three simple machines studied in this chapter. Using the work-energy theorem, explain why a simple machine does not reduce the total work done, only the effort. A person uses an inclined ramp of length 4 m to raise a box of mass 50 kg to a height of 1 m. Calculate the effort required and the mechanical advantage. Take g=10 m s−2.

Ans: Simple machines are devices that make work easier by changing the magnitude or direction of the force that needs to be applied. They form the building blocks of many everyday complex machines.

Three simple machines studied:

1. Pulley – a wheel with a groove that guides a rope to change the direction of the effort (e.g., used to raise flags or lift loads in construction).
2. Inclined Plane – a smooth sloping surface that allows a load to be raised or lowered to a height using a smaller effort applied over a longer distance (e.g., ramps, roads on hills).
3. Lever – a rigid bar that rotates about a fixed point (fulcrum), allowing a smaller effort to overcome a larger load or to change the direction of force (e.g., seesaw, scissors, crowbar).

Why total work done is not reduced:
By the work-energy theorem, the work done on the object equals the change in its energy. The potential energy gained by the load when raised to height $h$ is always $mgh$, regardless of the path taken or the machine used. Therefore:

$$\text{Effort}\times \text{Displacement of effort}=\text{Load}\times \text{Displacement of load}=mgh$$

A machine can reduce the effort required, but only by increasing the displacement over which that effort must be applied, so that the total work $(\text{effort}\times \text{displacement})$ remains the same. Machines do not create energy – they only help us use it more effectively by redistributing force and displacement.

Numerical Solution (Inclined Plane):

Load $=mg=50\times 10=500\text{ N}$, Height $h=1\text{ m}$, Length $L=4\text{ m}$

Using $F^{\mathrm{\prime }}\times L=mgh$:

$$F^{\mathrm{\prime }}=\frac{mgh}{L}=\frac{500\times 1}{4}=125\text{ N}$$

$$\text{MA}=\frac{L}{h}=\frac{4}{1}=4$$

The effort required is only 125 N (compared to 500 N for direct lifting), and the mechanical advantage is 4.

Ques 4: A jet aircraft of mass 15000 kg lands on an aircraft carrier. A wire exerts a constant backward force of 367500 N and stops the aircraft within 100 m. Using the work-energy theorem, find the velocity of the aircraft just before the wire caught the hook. Also explain what form of energy the kinetic energy gets converted into during braking.

Ans: Given: Mass $m=15000\text{ kg}$, Stopping force $F=367500\text{ N}$ (backward), Stopping distance $s=100\text{ m}$, Final velocity $v=0$.

Step 1 – Work done by the wire on the aircraft:
The wire force and displacement are in opposite directions, so work done is negative:

$W=F\times s=367500\times (-100)=-36750000\text{ J}$

Step 2 – Apply the work-energy theorem:

$$W=\mathrm{\Delta }K=K_f-K_i=0-\frac{1}{2}m{v}^2$$

$-36750000=-{\displaystyle \frac{1}{2}}\times 15000\times v^2$

$$v^2=\frac{36750000\times 2}{15000}=\frac{73500000}{15000}=4900{\text{ m}}^2{\text{s}}^{-2}$$

$$v=\sqrt{4900}=70{\text{ m s}}^{-1}=252{\text{ km h}}^{-1}$$

The velocity of the aircraft just before the wire caught the hook was $\mathbf{70}{\text{ m s}}^{-\mathbf{1}}$ (approximately 252 km h${}^{-1}$) towards the aircraft carrier.

Energy Conversion during braking:
The kinetic energy of the aircraft is converted primarily into:

1. Thermal energy (heat) – due to the stretching and tension in the wire and friction between the hook and wire.
2. Sound energy – due to the sudden deceleration and vibrations in the wire and aircraft structure.
3. Elastic potential energy – temporarily stored in the stretched wire before being released as heat and sound.

This is a direct application of the work-energy theorem: negative work done by the wire reduces the aircraft’s kinetic energy to zero.

Ques 5: Explain what is meant by elastic potential energy with two examples from daily life. A slingshot or an archer’s bow stores elastic potential energy. Explain how this energy is stored and then converted to kinetic energy of the projectile. Also explain the role of the work-energy theorem in understanding how a compressed spring launches an object.

Ans: Elastic potential energy is the energy stored in an object as a result of its deformation (change in shape or configuration) caused by an applied force. When the deforming force is removed, this stored energy is released and can do work on other objects. It is a type of potential energy.

Two examples from daily life:

1. Stretched rubber band (slingshot): When the rubber band of a slingshot is pulled back, an external force is applied to stretch it. Work is done against the internal (elastic) forces in the rubber, and this energy is stored as elastic potential energy. When released, the rubber returns to its original shape, doing work on the stone and converting the stored energy into kinetic energy of the stone.
2. Bent bow (archery): An archer applies force on the bowstring to bend the bow and stretch the string. Work is done against the internal elastic forces of the bow. This work gets stored as elastic potential energy in the deformed bow and string. When the archer releases the string, the bow and string return to their original positions, doing work on the arrow and transferring the stored elastic potential energy into kinetic energy of the arrow.

Role of work-energy theorem for a compressed spring:
When an external force compresses a spring, work is done on the spring against its internal restoring force. This work is stored as elastic potential energy $(U_s)$ in the spring. When the spring is released and allowed to expand:

1. The internal forces of the spring do positive work on the object in contact with it – the force is in the direction of the object’s displacement.
2. By the work-energy theorem: work done by the spring $=$ change in kinetic energy of the object:

$$U_s=\frac{1}{2}m{v}^2-0=\frac{1}{2}m{v}^2$$

The entire elastic potential energy stored in the spring is thus converted into kinetic energy of the launched object. The greater the compression (and therefore the greater the stored energy), the faster the object moves upon release. This shows that the work done on the spring during compression is fully recovered as kinetic energy of the projectile – a direct consequence of the work-energy theorem.