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Work, Energy, and Power physics ncert solutions class 11th Physics Class 11th Work, Energy and Power

6.13 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s^–1 ?

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Payal Gupta | Contributor-Level 10

5 months ago

6.13 The radius of the rain drop = 2 mm = 2 $* 10^{- 3}$ m

The height of drop, s = 500 m

Density of water, $?$ = $10^{3}$ kg/ $m^{3}$

Mass of the rain drop = volume $* d e n s i t y$ = (4/3) $? r^{3}$ $* ?$ = 3.35 $* 10^{- 5}$ kg

The gravitational force on the raindrop, F = mg = 3.28 $* 10^{- 4}$ N

Work done by the gravity on the drop is = mgs where s = 250 m

Work done = 0.082 J

The work done during the second half will remain same.

The total energy of the raindrop will be conserved during the motion.

Total energy at the top

E1 = mgh where h = 500 m, E1 = 0.164 J

Due to resistive force, the energy of the drop on reaching the ground

E2 = (1/2)mv² where

...more

6.13 The radius of the rain drop = 2 mm = 2 $* 10^{- 3}$ m

The height of drop, s = 500 m

Density of water, $?$ = $10^{3}$ kg/ $m^{3}$

Mass of the rain drop = volume $* d e n s i t y$ = (4/3) $? r^{3}$ $* ?$ = 3.35 $* 10^{- 5}$ kg

The gravitational force on the raindrop, F = mg = 3.28 $* 10^{- 4}$ N

Work done by the gravity on the drop is = mgs where s = 250 m

Work done = 0.082 J

The work done during the second half will remain same.

The total energy of the raindrop will be conserved during the motion.

Total energy at the top

E1 = mgh where h = 500 m, E1 = 0.164 J

Due to resistive force, the energy of the drop on reaching the ground

E2 = (1/2)mv² where v = 10 m/s, E2 = 0.001675 J

Work done by the resistive force, W = E1 – E2 = 0.1623 J

less

6.13 The radius of the rain drop = 2 mm = 2 <math><mo>*</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> mThe height of drop, s = 500 mDensity of water,  <math><mi>? </mi></math> = <math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> kg/ <math><msup><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Mass of the rain drop = volume <math><mo>*</mo><mi>d</mi><mi>e</mi><mi>n</mi><mi>s</mi><mi>i</mi><mi>t</mi><mi>y</mi></math> = (4/3) <math><mi>? </mi><msup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> <math><mo>*</mo><mi></mi><mi>? </mi></math> = 3.35 <math><mo>*</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>5</mn></mrow></mrow></msup></math> kgThe gravitational force on the raindrop, F = mg = 3.28 <math><mo>*</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>4</mn></mrow></mrow></msup></math> NWork done by the gravity on the drop is = mgs where s = 250 mWork done = 0.082 JThe work done during the second half will remain same.The total energy of the raindrop will be conserved during the motion.Total energy at the topE1 = mgh where h = 500 m, E1 = 0.164 JDue to resistive force, the energy of the drop on reaching the groundE2 = (1/2)mv2 where v = 10 m/s, E2 = 0.001675 JWork done by the resistive force, W = E1 – E2 = 0.1623 J

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Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

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This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

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A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

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(b) conserving energy between “O” ans ”A”

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$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

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...more

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv’

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v’)²=v²-2gh = v^’= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv’²]

1/2m(v₁)²=1/4m(v’)²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v’/ $\sqrt[]{2}$ =v²(v’/2)

v₁>v’/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’

(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.

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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

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This is a multiple choice answer as classified in NCERT Exemplar

Time of contact =0.001s

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