(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.For the first case,

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(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.For the first case,the body comes to rest from velocity $u$ after covering distance $s_{1}$:$F s_{1} = \frac{1}{2} m u^{2} \quad (i)$For the second case,the body comes to rest from velocity $2u$ after covering distance $s_{2}$:$F s_{2} = \frac{1}{2} m (2u)^{2} = \frac{1}{2} m (4u^{2}) = 4 \left( \frac{1}{2} m u^{2} \right) \quad (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{F s_{2}}{F s_{1}} = \frac{4 (\frac{1}{2} m u^{2})}{\frac{1}{2} m u^{2}}$$\frac{s_{2}}{s_{1}} = 4$$s_{2} = 4s_{1}$

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

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$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

KCET 2012 All KCET

A
$s_{2} = 4s_{1}$
B
$s_{2} = 2s_{1}$
C
$s_{2} = \frac{s_{1}}{2}$
D
$s_{2} = s_{1}$

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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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$A$ body of mass $m$ is travelling with a velocity $u$. When a constant retarding force $F$ is applied,it comes to rest after travelling a distance $s_{1}$. If the initial velocity is $2u$,with the same force $F$,the distance travelled before it comes to rest is $s_{2}$. Then,

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$A$ bullet is fired on a target with velocity $V$. Its velocity decreases from $V$ to $V/2$ when it penetrates $30 \text{ cm}$ in a target. Through what thickness will it penetrate further in the target before coming to rest (in $\text{ cm}$)?

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$A$ body of mass $10\, kg$ is released from a tower of height $20\, m$ and the body acquires a velocity of $10\, m/s$ after falling through the distance $20\, m$. The work done by the air resistance on the body is: ................. $J$ (Take $g = 10\, m/s^2$)

$A$ body of mass $2 \, kg$ is released from point $A$. Its velocity at point $B$ is $4 \, m/s$,and it comes to rest at point $C$. The work done against friction is ............. $J$.

The potential energy of a $1 \ kg$ particle free to move along the $x-$axis is given by: $U(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. The total mechanical energy of the particle is $2 \ J$. Then,the maximum speed (in $m/s$) is:

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