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New answer posted•a year ago

Work, Energy and Power Class 11th

In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point ‘B’ is

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

K_f – K_in = U_in - U_f

K - = mgy – mg (y – y₀) = mgy₀

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New answer posted•a year ago

Work, Energy and Power Class 11th

A car cover AB distance with first one-third at velocity $υ_{1} m s^{- 1},$ second one-third at $υ_{2} m s^{- 1}$ and last one-third at $υ_{3} m s^{- 1} .$ If  then the average velocity of the car is__________ ms^-1.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$t_{1} = \frac{L}{3 V_{1}}$

$t_{2} = \frac{L}{3 V_{2}}$

$t_{3} = \frac{L}{3 V_{3}}$

$V_{a m g} = \frac{L}{t_{1} + t_{2} + t_{3}} = \frac{L}{\frac{L}{3 V_{1}} + \frac{L}{3 V_{2}} + \frac{L}{3 V_{3}}}$

$V_{a m g} = \frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$

$= \frac{3}{\frac{1}{11} + \frac{1}{2 \times 11} + \frac{1}{3 \times 11}}$

= $\frac{3 \times 11 \times 6}{6 + 3 + 2}$

= 3 × b

= 18 m/sec

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

Water falls from a 40m height dam at the rate of 9 × 10⁴ kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is:

(Take g = 10 ms^-2)

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Given : water flow rate = $(\overset{o}{m})$ = 9 × 10⁴kg/hr

= $\frac{9 \times 1 0^{4} k g / s e c}{60 \times 60}$

Potential energy available on water per unit time

$\Rightarrow \overset{o}{m} g h = \frac{9 \times 1 0^{4}}{60 \times 60} \times 10 \times 40$

$\overset{o}{m} g h = 1 0^{4} w a t t$

50% of it converts into electrical energy = 50% of $\overset{o}{m} g h = \frac{1 0^{4}}{2} w a t t$

Let $η$ be the number of bulbs n = 100 = $\frac{1 0^{4}}{2}$

n = 50

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A block of mass 2kg moving on a horizontal surface with speed of 4 ms^-¹ enters a rough surface ranging from x = 0.5m to x = 1.5m. The retarding force in this range of rough surface is related to distance by F = -kx where k = 12Nm^-¹. The speed of the block as it just crosses the rough surface will be:

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

by, WET $\int F d x = Δ k . 6$

$\int - k x d x = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$\Rightarrow \frac{- k}{2} {[x_{f}^{2} - x_{i}^{2}]}_{0.5}^{1.5} = \frac{1}{2} \times 2 [v_{f}^{2} - 16]$ $[? v_{i} = 4 m / s e c]$

$\frac{- 12}{2} [1 . 5^{2} - 0 . 5^{2}] = [v_{f}^{2} - 16]$

$v_{f}^{2} = 16 - 12 = 4$

vf = 2m/sec

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New answer posted•a year ago

Work, Energy and Power physics ncert exemplar solutions class 12th chapter two

A system to 10 balls each of mass 2kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth in figure. Tension on the string between the 4^th and 8^th ball is_________N when 6^th ball just leaves the table.

20 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

12 g – T = 12 a- (1)

T – T’ = 2a- (2)

T’ = 6a- (3)

From (1) & (2)

12g – 12a – T’ = 2a- (4)

From (4) & (3)

12g – 12a – T’ = 14

$12 g = \frac{10}{3} T' \Rightarrow T' = \frac{120 \times 3}{10} = 36$

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A fighter jet is flying horizontally at a certain with a speed of 200ms^-¹. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle q with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of q will be _____________.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

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New answer posted•a year ago

Work, Energy and Power physics ncert exemplar solutions class 12th chapter two

Arrange the four graphs in descending order of total work done; where W₁, W₂, W₃ and W₄ are the work done corresponding to figure a, b, c and d respectively.

(a) W₃ > W₂ > W₁ >W₄ (b) W₃ > W₂ > W₄ > W₁ (c)  (d) W₂> W₃ > W₁ > W₄

34 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Work done represents are under

F vs x diagram

w₃ > w₂ > w₁ > w₄

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A particle of mass 500 gm is moving in a straight line with velocity =  The work done by the net force during its displacement from x = 0 to x = 4m is:

(Take b = 0.25m^-3/2s^-1).

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$ω = Δ k . E = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$= \frac{1}{2} m {[b \times 4^{5 / 2}]}^{2} - \frac{1}{2} m {[0]}^{2}$

$= \frac{1}{2} \times 0.5 [0.2 5^{2} \times 4^{5}] = 2^{4} = 16 J$

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Six

A body mass 8kg and another of mass 2kg are moving with equal kinetic energy. Thee ratio of their respective momenta will be:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$K E = \frac{ρ^{2}}{2 m}$

$\Rightarrow ρ^{2} α m$

$\Rightarrow {(\frac{P_{1}}{P_{2}})}^{2} = \frac{8}{2} = \frac{4}{1}$

$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{2}{1}$

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New answer posted•a year ago

Work, Energy and Power Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

A bullet of mass 200g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If it’s kinetic energy reduces to 40 J within 1s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

m = 200 g, ki = 90 J kf = 40 J., $Δ t = 1 s e c$

$\Rightarrow F = \frac{k_{f} - k_{i}}{x} \Rightarrow \sqrt[]{. 4 \times 90} - \sqrt[]{. 4 \times 40} = F \times 1$

$\Rightarrow F = 6 - 4 = 2 N$

So for complete rest :-

Fx = 90 J

2x = 90 x = 45 m

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