Work, Energy and Power

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

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Contributor-Level 10

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

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

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

Work done $= Δ K$

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

=> W = 16 J

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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

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Contributor-Level 10

Work done = $Δ K E$

$\Rightarrow - f = - x$

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

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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

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Contributor-Level 10

Elastic potential energy stored in the spring

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

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

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

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

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

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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

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Contributor-Level 10

Height attained after first collision (h1) = e²h0

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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)

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Contributor-Level 10

Let v is velocity at the highest position.

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

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

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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

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Contributor-Level 10

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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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)

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Raj Pandey

Contributor-Level 9

From conservation of Energy:

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

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

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

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A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration (a) is varying with time t as a = k² rt², where k is a constant. The power delivered to the particle by the force acting on it is given as

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Raj Pandey

Contributor-Level 9

a = k2rt2

$\Rightarrow \frac{v^{2}}{r} = k^{2} r t^{2}$

$\Rightarrow v = k r t$

$a_{t} = \frac{d v}{d t} = k r$

? tangential force, F_t = ma_t = mkr

$∴ P o w e r d e l i v e r e d, P = F_{t} v = (m k r) (k r t) = m k^{2} r^{2} t$

Note ® Power delivered by centripetal force will be zero.

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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

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