(D) Using the law of conservation of energy between point $A$ and point $B$:$E_A = E_B$$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$$v_B = \sqrt{v_A^2

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(D) Using the law of conservation of energy between point $A$ and point $B$:$E_A = E_B$$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$$v_B = \sqrt{v_A^2 + 2gh}$Given $v_A = 5\,m/s$,$g = 10\,m/s^2$,and $h = 10\,m$:$v_B = \sqrt{5^2 + 2 \times 10 \times 10} = \sqrt{25 + 200} = \sqrt{225} = 15\,m/s$At point $B$,the particle is at a distance $r = a = 10\,m$ from point $O$. The velocity vector at $B$ is perpendicular to the radius $OB$.The angular momentum $L$ about $O$ is given by:$L = m \cdot v_B \cdot r$$L = (20 \times 10^{-3}\,kg) \times (15\,m/s) \times (10\,m)$$L = 0.02 \times 150 = 3\,kg \cdot m^2/s$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Medium

$A$ particle of mass $20\,g$ is released with an initial velocity $5\,m/s$ along the curve from the point $A,$ as shown in the figure. The point $A$ is at height $h = 10\,m$ from point $B.$ The particle slides along the frictionless surface. When the particle reaches point $B,$ its angular momentum about $O$ will be ......... $kg \cdot m^2/s$. [Take $g = 10\,m/s^2$]

JEE MAIN 2019 All JEE MAIN

A
$2$
B
$8$
C
$6$
D
$3$

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Class 11

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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Derive the work-energy theorem for a variable force exerted on a body in one dimension.

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$A$ mass of $2 \ kg$, initially at a height of $1.2 \ m$ above an uncompressed spring with spring constant $2 \times 10^4 \ N/m$, is released from rest to fall on the spring. Taking the acceleration due to gravity as $10 \ m/s^2$ and neglecting air resistance, the compression of the spring in $mm$ is:

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The system shown in the figure is released from rest. The pulley and spring are massless,and friction is absent everywhere. The speed of the $5 \, kg$ block when the $2 \, kg$ block leaves contact with the ground is (take the force constant of the spring $k = 40 \, N/m$ and $g = 10 \, m/s^2$):

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$A$ body of mass $M$ is released from a height of $1.5 \ m$ on a curved track. It slides down and moves along a horizontal surface $PQ$ of length $15 \ m$. If the coefficient of friction between the body and the horizontal surface is $0.2$,at what distance from $P$ will the body come to rest (in $m$)?

Medium

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Potential energy of a body of mass $1 \ kg$ free to move along $X$-axis is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the body is $2 \ J$,then the maximum speed of the body is (Assume only conservative force acts on the body).

MediumAP EAMCET 2019

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Derive the work-energy theorem for a variable force exerted on a body in one dimension.

The system shown in the figure is released from rest. The pulley and spring are massless,and friction is absent everywhere. The speed of the $5 \, kg$ block when the $2 \, kg$ block leaves contact with the ground is (take the force constant of the spring $k = 40 \, N/m$ and $g = 10 \, m/s^2$):

Potential energy of a body of mass $1 \ kg$ free to move along $X$-axis is given by $U(x) = (\frac{x^2}{2} - x) \ J$. If the total mechanical energy of the body is $2 \ J$,then the maximum speed of the body is (Assume only conservative force acts on the body).

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