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Work, Energy and Power

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2 months ago

Work, Energy and Power Class 11th

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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2 months 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

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alok kumar singh

Contributor-Level 10

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

$\Rightarrow$ K_f – K_in = U_in - U_f

$\Rightarrow$ K - = mgy – mg (y – y₀) = mgy₀

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2 months ago

Work, Energy and Power Class 11th

A 1kg particle is moving unidirectionally on a horizontal plane under the action of power $P = - 2 v^{3}$ , of power supplying energy source. The velocity (v) - time (t) graph that describes the motion of the particle is shown in figure. (Graphs are drawn schematically and are not to scale) The value of $t_{0}$ in graph is sec. Find 8x

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Raj Pandey

Contributor-Level 9

Kindly consider the following Image

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2 months ago

Work, Energy and Power Class 11th

A particle of mass 2 kg is moving in the direction of positive x-axis. Initial velocity of the particle is 1 m/s. The instantaneous power vs time graph of the particle is shown in the figure. Find the velocity of the particle at 7 seconds.

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Vishal Baghel

Contributor-Level 10

ΔW = area under P — t graph
= (1/2) (4 + 6) * 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) * 2 * v² - (1/2) * 2 * (1)² ⇒ v = 6 m/s

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2 months ago

Work, Energy and Power Class 11th

In an imaginary electric field, the electric potential energy of an electron at a distance r from the centre of force field is given by U = kr², where k is a positive constant of appropriate dimensions. If the electron is moving in a circular orbit of radius r about the centre, then the orbital time period of the electron is proportional to:

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Vishal Baghel

Contributor-Level 10

U = kr² ⇒ F = -dU/dr = -2kr ; 2kr = mv²/r ⇒ v = √ (2k/m) or T = 2πr/v = 2π√ (m/2k)

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2 months ago

Work, Energy and Power Class 11th

A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)

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Raj Pandey

Contributor-Level 9

Vertical component of velocity just after collision $= \frac{u}{2 \sqrt[]{2}}$

$k_{i} = \frac{m u^{2}}{2} k_{f} = \frac{1}{2} m \frac{u^{2}}{2} + \frac{1}{2} m \frac{u^{2}}{8} = \frac{5 m u^{2}}{16}$ $\frac{}{}$

Fraction $= \frac{k_{i} - k_{f}}{k_{i}} = 1 - \frac{\frac{5 m u^{2}}{16}}{\frac{m u^{2}}{2}} = 1 - \frac{5}{8} = \frac{3}{8}$

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2 months ago

Work, Energy and Power Class 11th

What minimum speed does a 100g particle need At point B to reach point A? The graph shows potential energy versus position.

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alok kumar singh

Contributor-Level 10

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

$\frac{1}{2} * \frac{100}{1000} * v^{2} = (5 - 0)$

v² = 100

v = 10 m/s

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2 months ago

Work, Energy and Power Class 11th

Figure shown two blocks A and B having mass 2kg and 4kg moving with a speed 4 m/sec and 2 m/sec respectively. The maximum compression in the spring and final velocity of 2kg block and respectively.

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alok kumar singh

Contributor-Level 10

$\frac{1}{2} μ v_{r}^{2} = \frac{1}{2} k x^{2}$

$x = {V_{r e r}}^{2} \sqrt[]{\frac{μ}{k}}$

$= 2 * \sqrt[]{\frac{8 / 6}{10}} = \sqrt[]{\frac{8}{15}}$

From conservation of momentum

2 * 4 + 4 * 2 = 2v₁ + 4v₂

8 = v₁ + 2v₂ ….(1)

From conservation of energy

$\frac{1}{2} * 2 * 4^{2} + \frac{1}{2} * 4 * 2^{2} = \frac{1}{2} * 2 * v_{1}^{2} + \frac{1}{2} * 4 * v_{2}^{2}$ ….(ii)

On solving $v_{1} = \frac{4}{3} m / s, v_{2} = \frac{10}{3} m / s$

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2 months ago

Work, Energy and Power Class 11th

A plank of mass 5 kg in kept on a smooth surface. A block of 1 kg is kept on it and is given a velcoity of 12 m/s. There is friction between the block and the plank. Assuming the plank to be very long, find the magnitude of work done (in Joules) by friction on block till the relative sliding between them stops.

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Vishal Baghel

Contributor-Level 10

When relative sliding stops, both move with same velocity.
Only friction acts in horizontal direction.
∴ for system no external force, we can apply conservation of momentum.
1 * 12 + 5 * 0 = 6 * v
v = 2 m/s
Now work done by friction on block = change in kinetic energy of block
w = (1/2)m (2² - 12²) = (1/2) (1) (4 - 144) = (1/2) (-140) = -70 J

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2 months ago

Work, Energy and Power Class 11th

An inclined plane in bent in such a way that the vertical cross-section is given by y $= \frac{x^{2}}{4}$ where y is in vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coefficient of friction $μ$ = 0.5, the maximum height in cm at with a stationary block with not skip downward is ____________cm.

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alok kumar singh

Contributor-Level 10

If the block does not slide,

mg sin $θ \leq μ m g c o s θ$

$\Rightarrow t a n θ \leq μ \Rightarrow \frac{d y}{d x} \leq μ \Rightarrow \frac{x}{2} \leq 0.5 \Rightarrow x \leq 1 m .$

Thus, $y \leq \frac{1^{2}}{4} = 0.25 m = 25 c m .$

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