Work, Energy and Power

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New answer posted•a year ago

Work, Energy and Power Class 11th

Two springs with spring constants $k_{1} = 1200 N / m$ and $k_{2} = 500 N / m$ are stretched by the same force. The ratio of potential energy in the springs will be

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Elastic potential energy stored in the spring

$U = \frac{1}{2} k x^{2}$

$= \frac{1}{2} F x = \frac{F^{2}}{2 k}$

$H e n c e \frac{U_{1}}{U_{2}} = \frac{k_{2}}{k_{1}} [F i s s a m e f o r b o t h s p r i n g s]$

$= \frac{500}{1200} = \frac{5}{12}$

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A mass of $0.5 k g$ moving with a speed of $1 . 5 m / s$ on a horizontal smooth surface, collides with a nearly weightless spring of force constant $k = 50 N / m$ . The maximum compression of the spring would be

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$M = 0.5 kg$

$V = 1.5 m / s$

$K = 50 N / m$

Maximum Compression of the spring

$\frac{1}{2} m v^{2} = \frac{1}{2} k x^{2}$

$k x^{2} = m v^{2}$
$x = v \sqrt[]{\frac{m}{k}} = 1.5 \sqrt[]{\frac{0.5}{50}} .$

$x = 0.15 m$

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

The position x of a particle of mass 2kg moving along straight line is given as x = (t – 2)², where t is in second and x is in metre. The work done by the force in first 2s is

19 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} \Rightarrow x = {(t - 2)}^{2} \\ \Rightarrow v = 2 t - 4 \end{array}$

$v_{t = 0} = - 4 m / s$

$v_{t = 2} = 0 m / s$

Work done $= Δ K$

$\Rightarrow W = \frac{1}{2} m \times 0 - \frac{1}{2} \times 2 \times 16$

=> W = 16 J

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A body possesses kinetic energy x, moving on a rough horizontal surface, is stopped in a distance y. The friction force exerted on the body is

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Work done = $Δ K E$

$\Rightarrow - f = - x$

$\Rightarrow f = \frac{x}{y}$

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

Two springs with spring constants $k_{1} = 1200 N / m$ and $k_{2} = 500 N / m$ are stretched by the same force. The ratio of potential energy in the springs will be

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Elastic potential energy stored in the spring

$U = \frac{1}{2} k x^{2}$

$= \frac{1}{2} F x = \frac{F^{2}}{2 k}$

$H e n c e \frac{U_{1}}{U_{2}} = \frac{k_{2}}{k_{1}} [F i s s a m e f o r b o t h s p r i n g s]$

$= \frac{500}{1200} = \frac{5}{12}$

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

The kinetic energy of a body decreases by 36%. The decrease in linear momentum is

9 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A particle of mass m is dropped from rest from a height h0 on a horizontal floor. The coefficient of restitution between the floor and particle is e. Maximum height attained by the particle after second collision is

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Height attained after first collision (h1) = e²h0

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……………… m/s. (Take g = 10 m/s²)

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Let v is velocity at the highest position.

$T_{m a x} = 5 T_{m i n} \Rightarrow m g + \frac{m (v^{2} + 4 g l)}{l} = 5 (\frac{m v^{2}}{l} - m g) \Rightarrow 4 . \frac{v^{2}}{l} = 10 g$

$\Rightarrow v = \sqrt[]{\frac{5}{2} g l} = \sqrt[]{\frac{5}{2} \times 10 \times 1} = 5 m / s$

Upvote

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

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | 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

Upvote

New answer posted•a year ago

Work, Energy and Power Class 11th

A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200g. The bob is pulled aside until the string is at 60° with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be_________ ms^-¹. (if g = 10m/s²)

14 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

From conservation of Energy:

$m g l (1 - c o s 60 °) = \frac{1}{2} m v^{2}$

$\Rightarrow v^{2} = 2 g l (\frac{1}{2}) = g l$

$= 10 \times \frac{25}{10} = 5 m / s$

Upvote

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