(B) According to the work-energy theorem,the loss in potential energy is equal to the work done against friction on the horizontal surface.Let $h = 1.

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(B) According to the work-energy theorem,the loss in potential energy is equal to the work done against friction on the horizontal surface.Let $h = 1.5 \ m$ be the initial height and $\mu = 0.2$ be the coefficient of friction.The potential energy at the start is $PE = Mgh$.The work done by friction over a distance $s$ on the horizontal surface is $W_f = f \cdot s = (\mu Mg) \cdot s$.Equating the two: $Mgh = \mu Mgs$.Solving for $s$: $s = \frac{h}{\mu} = \frac{1.5}{0.2} = 7.5 \ m$.Since $7.5 \ m < 15 \ m$,the body comes to rest at a distance of $7.5 \ m$ from point $P$.

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

Medium

$A$ 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 $m$)?

A
$1$
B
$7.5$
C
$2$
D
$5$

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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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$A$ body of mass $5 \ kg$ is placed on a horizontal surface with a coefficient of friction $\mu = 0.2$. It is pulled by a horizontal force of $25 \ N$ for a distance of $10 \ m$. The kinetic energy gained by the body is ..... $J$. (Take $g = 10 \ m/s^2$)

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$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

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