Chapter 7: Work, Energy, and Simple Machines

If friction is ignored, the potential energy of the child at the top of the slide changes into kinetic energy at the bottom.

At the top:

PE = mgh

At the bottom:

KE = ½mv²

By conservation of mechanical energy:

mgh = ½mv²

v = √(2gh)

So, the magnitude of velocity of the child at the bottom of the blue slide will be:

v = √(2gh)

where h is the vertical height of the slide.

Yes, if friction is ignored, two children of different masses will reach the bottom of the same slide with the same velocity.

This is because:

v = √(2gh)

The formula does not contain mass. So, the velocity depends only on the height of the slide and acceleration due to gravity, not on the mass of the child.

The slide that has the greatest vertical height will result in the largest magnitude of velocity at the bottom.

This is because:

v = √(2gh)

Greater height means greater potential energy at the top, which changes into greater kinetic energy at the bottom. If all slides start from the same vertical height, then the child will reach the bottom with the same velocity, ignoring friction.

No, the weightlifter is not doing any work on the barbell in the scientific sense while holding it steady.

Work is done only when force causes displacement:

W = F × s

Here, the weightlifter applies force on the barbell, but the barbell does not move.

So,

s = 0

Therefore,

W = F × 0 = 0

Thus, work done on the barbell is zero. However, the weightlifter may feel tired because her muscles use internal energy while holding the barbell.

The work done by friction on the stack of coins is negative.

This is because friction acts opposite to the direction of motion of the coins. When force and displacement are in opposite directions, work done is negative.

So, friction does negative work and reduces the kinetic energy of the coins.

When we pedal a bicycle on a flat road, the chemical energy stored in our muscles changes into different forms:

• Kinetic energy of the bicycle and rider because they move forward.

• Thermal energy due to friction between tyres and road, chain and gears, and air resistance.

• Sound energy in the form of small sounds produced by the bicycle.

Since the road is flat, there is no significant increase in gravitational potential energy.

Let the velocity of object A be v_A , and the velocity of object B be v_B .

Kinetic energy:

K = ½mv²

For object A:

K_A = ½mv_A²

For object B:

K_B = ½(4m)v_B²

Since both have the same kinetic energy:

½mv_A² = ½(4m)v_B²

v_A² = 4v_B²

v_A = 2v_B

Therefore,

v_A:v_B = 2:1

So, the lighter object A moves with twice the velocity of object B.

No, the kinetic energy of an object moving with constant velocity does not change with its position.

Kinetic energy is given by:

K = ½mv²

It depends on mass and velocity. If the mass and velocity remain constant, kinetic energy also remains constant, even if the object changes its position.

The gravitational potential energy near the surface of the Earth is:

U = mgh

If an object moves horizontally with constant velocity, its height does not change. So, its potential energy does not change.

But if the object is gradually raised in the vertical direction, its height increases. Therefore, its potential energy increases.

So:

• Horizontal motion at same height → potential energy remains same.

• Vertical upward motion → potential energy increases.

At the top, the ball is at height h .

Potential energy at the top:

PE = mgh

Kinetic energy at the top:

KE = 0

Total mechanical energy at the top:

ME = PE + KE = mgh + 0 = mgh

Just before the ball hits the ground:

PE = 0

By conservation of mechanical energy, the lost potential energy changes into kinetic energy.

So,

KE = mgh

Therefore, mechanical energy just before hitting the ground is:

ME = KE + PE

ME = mgh + 0 = mgh

Hence, even just before hitting the ground, the mechanical energy of the ball is:

mgh

At point A, the ball is at the highest point. So, its potential energy is maximum and kinetic energy is minimum or zero if it is released from rest.

At point B, the ball is at a lower position. Its potential energy decreases and changes into kinetic energy. So, kinetic energy becomes maximum near the lowest point.

At point C, the ball rises again. Its kinetic energy decreases and changes back into potential energy.

The later points such as C, D and E usually have lower heights because some mechanical energy is lost due to friction and air resistance. This lost energy changes into heat and sound. Therefore, the ball cannot rise to the same height again.

Yes, this lowering of height is due to energy lost because of friction and air resistance.

Roads on hills are built to wind around in gentle slopes because they act like inclined planes.

A gentle slope increases the distance travelled but reduces the force required to move upward. If the road went straight up, vehicles would need a much larger force to climb.

So, hill roads are made longer and less steep to make climbing easier and safer for vehicles.

An inclined ladder is easier to climb because it works like an inclined plane.

In an inclined ladder, we move through a larger distance, but the effort required at each step is less. In a vertical ladder, we have to lift our body almost directly against gravity, so it requires more effort.

Therefore, climbing an inclined ladder feels easier than climbing a vertical ladder.

It is easier to open the lid of a can using a spoon because the spoon acts as a lever.

The edge of the can acts as the fulcrum, the lid is the load, and the force applied by our hand is the effort. Since the effort arm is longer, a small force applied at the handle of the spoon produces a larger force at the lid.

Thus, the spoon gives mechanical advantage and makes opening the lid easier.

We push a hard object closer to the fulcrum of the scissors because the cutting force is greater near the fulcrum.

In scissors, the fulcrum is the screw or pivot. When the object is closer to the fulcrum, the load arm becomes smaller. This increases the mechanical advantage of the scissors.

So, a larger force acts on the hard object and it becomes easier to cut.

All real machines eventually slow down and stop because resistive forces like friction and air resistance act on them.

These forces do negative work on the machine. As a result, the mechanical energy of the machine gradually decreases. This energy is not destroyed, but it changes into other forms such as heat and sound.

A machine cannot create energy on its own. To continue doing useful work forever, it would need a continuous supply of energy. Therefore, perpetual motion machines do not work in real life.

Machines can only transfer or convert energy; they cannot create unlimited energy.

False

Explanation: Work is done only when force produces displacement. If the object does not move, displacement is zero, so work done is zero.

True

Explanation: While lifting the bucket, the applied force and displacement are both upward. Therefore, positive work is done on the bucket.

True

Explanation: The SI unit of both work and energy is joule (J).

False

Explanation: A motionless stretched rubber band does not have kinetic energy. It has elastic potential energy due to its stretched shape.

True

Explanation: Energy can change from one form to another, for example, electrical energy changes into light energy in a bulb.

Work done = Force × Displacement in the direction of force.

1 joule of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.

K = ½mv²

U = mgh

Power is defined as the rate at which work is done.

Correct statements are:

(iii) Its kinetic energy is zero.

(iv) Its potential energy is maximum.

Explanation:

At the highest point, the velocity of the ball becomes zero for an instant. Therefore, its kinetic energy is zero.

However, gravitational force still acts downward on the ball, so force is not zero. Acceleration due to gravity also acts downward, so acceleration is not zero. Since the ball is at maximum height, its potential energy is maximum.

The required energy transformations are shown in the table below.

Given:

m = 50 kg

g = 10 m s⁻²

h = 72.5 m

Potential energy gained:

U = mgh

U = 50 × 10 × 72.5

U = 36250 J

(i) Gain in potential energy when lifted straight up:

36250 J

(ii) Gain in potential energy when the student climbs the stairs:

36250 J

(iii) Conclusion:

The gain in potential energy depends only on the mass, acceleration due to gravity, and vertical height gained. It does not depend on the path taken.

So, whether the student is lifted straight up or climbs the stairs, the gain in potential energy remains the same.

Let the height of the 10th floor be h . Then the height of the 20th floor will be 2h .

Energy required to lift the mass to the 10th floor:

E₁ = mgh

Energy required to lift the same mass to the 20th floor:

E₂ = mg(2h) = 2mgh

So, the energy required is double.

Now, let the time taken to lift the mass to the 10th floor be t . Time taken to lift it to the 20th floor is 2t .

Power for 10th floor:

P₁ = mgh/t

Power for 20th floor:

P₂ = 2mgh/2t

P₂ = mgh/t

So,

P₂ = P₁

Final Answer:

The crane requires twice the energy, but the same power, because the time taken is also doubled.

The energy required to raise a flag depends mainly on:

• Mass of the flag

• Height of the flagpole

• Acceleration due to gravity

• Friction in the pulley, if present

Ignoring friction, work done or energy required is:

W = mgh

Raising the flag slowly or quickly does not change the amount of work done, because the height and weight of the flag remain the same.

But power depends on time:

P = W/t

If the speed of raising the flag is doubled, the same work is done in half the time. Therefore, the power requirement becomes double.

Final Answer:

The work done remains the same, but if the speed is doubled, the power required is doubled.

Mass of scooter = 100 kg Mass of man = 60 kg

Total mass on first day:

100 + 60 = 160 kg

Mass of son = 40 kg

Total mass on second day:

100 + 60 + 40 = 200 kg

Kinetic energy required:

K = ½mv²

Since velocity is the same on both days, kinetic energy depends only on total mass.

Ratio of fuel used:

First day : Second day = 160:200

= 4:5

Final Answer:

4:5

So, the fuel used on the first day and second day is in the ratio 4 : 5.

For a balanced seesaw:

Weight of child × Distance of child from fulcrum = Weight of adult × Distance of adult from fulcrum

Let the weight of the child be W . Then the weight of the adult is 2W .

W × d_c = 2W × d_a

d_c = 2d_a

So, the child should sit twice as far from the fulcrum as the adult.

Example:

Child Fulcrum Adult

W ▲ 2W

| | |

|---------- 2 m ---------------|------- 1 m --------|

Final Answer:

The lighter child should sit farther from the fulcrum, and the heavier adult should sit closer to the fulcrum. If the adult sits 1 m from the fulcrum, the child should sit 2 m from the fulcrum.

Given:

m = 2 kg

u = 20 m s⁻¹

g = 10 m s⁻²

h = 19.4 m

(i) Sign of work done by gravity

During upward motion, displacement is upward but gravity acts downward. So, work done by gravity is negative.

During downward motion, displacement is downward and gravity also acts downward. So, work done by gravity is positive.

Upward motion: Negative work Downward motion: Positive work

(ii) Work done by air resistance

Initial kinetic energy of the ball:

K = ½mv²

K = ½ × 2 × 20²

K = 400 J

At the highest point, velocity becomes zero, so kinetic energy is zero.

Potential energy at height 19.4 m:

U = mgh

U = 2 × 10 × 19.4

U = 388 J

Initial mechanical energy = 400 J Final mechanical energy = 388 J

Work done by air resistance:

W = 388 - 400

W = - 12 J

Final Answer:

- 12 J

The negative sign shows that air resistance does negative work and reduces the mechanical energy of the ball.

Given:

m = 10.0 kg

Initial kinetic energy at 0 m:

K_i = 180 J

(i) Speed at 0 m

K = ½mv²

180 = ½ × 10 × v²

180 = 5v²

v² = 36

v = 6 m s⁻¹

So, the speed at 0 m is:

6 m s⁻¹

Work done from 0 m to 4 m

Work done is equal to the area under the force-displacement graph.

From the graph:

Triangle from 0 m to 1 m:

W₁ = ½ × 1 × 50 = 25 J

Rectangle from 1 m to 3 m:

W₂ = 2 × 50 = 100 J

Triangle from 3 m to 4 m:

W₃ = ½ × 1 × 50 = 25 J

Total work done:

W = 25 + 100 + 25

W = 150 J

By work-energy theorem:

Work done = Change in kinetic energy

W = K_f - K_i

150 = K_f - 180

K_f = 330 J

(ii) Speed at 4 m

K_f = ½mv²

330 = ½ × 10 × v²

330 = 5v²

v² = 66

v = √(66)

v ≈ 8.12 m s⁻¹

So, the speed at 4 m is:

8.12 m s⁻¹

Does the block have negative acceleration?

No, the block does not have negative acceleration in any portion of its motion.

The force is always in the direction of motion or zero. Since acceleration depends on force, the acceleration is positive where force is positive and zero where force is zero. It is never negative.

On Earth, maximum height:

h_E = 8 m

The gravitational acceleration on the Moon is:

g_M = g_E/6

For the same initial upward velocity:

h = u²/2g

Height is inversely proportional to gravitational acceleration.

So,

h_M = 6h_E

h_M = 6 × 8

h_M = 48 m

Therefore, the ball will rise up to:

48 m

on the Moon.

Between A and B, the car moves with a constant speed of 35 m s⁻¹.

This means the car is moving uniformly during this interval. The driver has noticed the obstruction, but braking has not yet reduced the speed.

Given:

m = 1000 kg

v = 35 m s⁻¹

K = ½mv²

K = ½ × 1000 × 35²

K = 500 × 1225

K = 612500 J

So, kinetic energy at A is:

612500 J

or

6.125 × 10⁵ J

At B, kinetic energy is:

612500 J

At C, the car stops, so kinetic energy is:

0 J

Work done by brakes:

W = K_f - K_i

W = 0 - 612500

W = - 612500 J

So, the work done by the brakes is:

- 612500 J

The negative sign shows that the braking force acts opposite to the motion of the car.

The kinetic energy of the car mainly transforms into thermal energy due to friction in the brakes, tyres and road. A small part may also change into sound energy.

Given:

m = 0.5 kg

At O:

v = 0

PE = 30 J

So, kinetic energy at O is:

KE = 0

Total mechanical energy:

ME = PE + KE

ME = 30 + 0 = 30 J

Since the track is frictionless, total mechanical energy remains constant.

ME = 30 J

At P

Potential energy at P:

PE_P = 20 J

KE_P = ME - PE_P

KE_P = 30 - 20 = 10 J

Now,

KE = ½mv²

10 = ½ × 0.5 × v²

10 = 0.25v²

v² = 40

v = √(40)

v ≈ 6.32 m s⁻¹

So, velocity at P is:

6.32 m s⁻¹

At Q

Potential energy at Q:

PE_Q = 30 J

KE_Q = ME - PE_Q

KE_Q = 30 - 30 = 0 J

So,

v_Q = 0 m s⁻¹

Velocity at Q is:

0 m s⁻¹

At R

Potential energy at R:

PE_R = 40 J

But total mechanical energy of the ball is only:

ME = 30 J

Since the potential energy at R is greater than the total mechanical energy, the ball cannot reach point R on its own.

So, velocity at R is:

The ball cannot reach R

Given:

m = 1.5 kg

h = 10 m

g = 10 m s⁻²

(i) Velocity just before hitting the sand

Using conservation of mechanical energy:

mgh = ½mv²

Mass cancels out:

gh = ½v²

v² = 2gh

v = √(2gh)

v = √(2 × 10 × 10)

v = √(200)

v ≈ 14.14 m s⁻¹

So, the velocity of the coconut just before hitting the sand is:

14.14 m s⁻¹

downward.

(ii) Depth of depression in the sand

Potential energy of coconut at the top:

PE = mgh

PE = 1.5 × 10 × 10

PE = 150 J

This energy is used to create depression in the sand.

Work done against resistive force of sand:

W = F × d

Given:

F = 3000 N

W = 150 J

150 = 3000 × d

d = 150/3000

d = 0.05 m

0.05 m = 5 cm

So, the depth of depression is:

0.05 m

or

5 cm