(A) The rate of doing work is power,$P = Fv$. Given $P = 2x$.Since $F = ma = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt} = mv \frac{dv}{dx}$,we ha

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$\left[\frac{3x^2}{m}\right]^{1/3}$

(A) The rate of doing work is power,$P = Fv$. Given $P = 2x$.Since $F = ma = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt} = mv \frac{dv}{dx}$,we have:$P = (mv \frac{dv}{dx}) v = mv^2 \frac{dv}{dx} = 2x$.Rearranging the terms for integration:$mv^2 dv = 2x dx$.Integrating both sides:$\int mv^2 dv = \int 2x dx$.$\frac{mv^3}{3} = x^2 + C$.Assuming the particle starts from rest at the origin ($v=0$ at $x=0$),the constant $C = 0$.Therefore,$\frac{mv^3}{3} = x^2$.$v^3 = \frac{3x^2}{m}$.$v = \left[\frac{3x^2}{m}\right]^{1/3}$.

Class 11 Physics Work, Energy, Power and Collision Mix Examples-Work, Energy, Power and Collision

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The rate of doing work by a force acting on a particle moving along the $x$-axis depends on the position $x$ of the particle and is equal to $2x$. The velocity of the particle is given by the expression:

A
$\left[\frac{3x^2}{m}\right]^{1/3}$
B
$\left[\frac{3x^2}{2m}\right]^{1/3}$
C
$\left(\frac{2mx}{9}\right)^{1/2}$
D
$\left[\frac{mx^2}{3}\right]^{1/2}$

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Work, Energy, Power and Collision

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Mix Examples-Work, Energy, Power and Collision

$A$ section of a fixed smooth circular track of radius $R$ in a vertical plane is shown in the figure. $A$ block is released from position $A$ and leaves the track at $B$. The radius of curvature of its trajectory when it just leaves the track at $B$ is:

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$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. The maximum height $h$ attained by the particle is:

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Two trolleys of mass $m$ and $3m$ are connected by a spring. They are compressed and released; once released,they move off in opposite directions and come to rest after covering distances $S_1$ and $S_2$ respectively. Assuming the coefficient of friction to be uniform,the ratio of distances $S_1:S_2$ is:

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$A$ block of mass $1 \ kg$ is moved by a distance of $1 \ m$ on a horizontal surface by applying a force of $8 \ N$. Then it is lifted to a height of $2 \ m$. The total work done is ... $J$. (Take $g = 10 \ m/s^2$)

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$A$ $0.5\, kg$ ball is thrown up with an initial speed $14\, m/s$ and reaches a maximum height of $8.0\, m$. How much energy is dissipated by air drag acting on the ball during the ascent? (Take $g = 9.8\, m/s^2$)

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The rate of doing work by a force acting on a particle moving along the $x$-axis depends on the position $x$ of the particle and is equal to $2x$. The velocity of the particle is given by the expression:

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$A$ section of a fixed smooth circular track of radius $R$ in a vertical plane is shown in the figure. $A$ block is released from position $A$ and leaves the track at $B$. The radius of curvature of its trajectory when it just leaves the track at $B$ is:

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. The maximum height $h$ attained by the particle is:

$A$ block of mass $1 \ kg$ is moved by a distance of $1 \ m$ on a horizontal surface by applying a force of $8 \ N$. Then it is lifted to a height of $2 \ m$. The total work done is ... $J$. (Take $g = 10 \ m/s^2$)

$A$ $0.5\, kg$ ball is thrown up with an initial speed $14\, m/s$ and reaches a maximum height of $8.0\, m$. How much energy is dissipated by air drag acting on the ball during the ascent? (Take $g = 9.8\, m/s^2$)

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