(D) According to the Work-Energy Theorem,the total work done on the block is equal to the change in its kinetic energy: $W_{gravity} + W_{friction} =

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(D) According to the Work-Energy Theorem,the total work done on the block is equal to the change in its kinetic energy: $W_{gravity} + W_{friction} = \Delta K$.Since the block starts from rest and comes to rest on the horizontal surface,$\Delta K = 0$.The work done by gravity is $W_g = mgh$.The work done by friction is $W_f = -f \cdot S = -\mu N \cdot S = -\mu mg \cdot S$.Setting the sum to zero: $mgh - \mu mgS = 0$.Dividing by $mg$: $h - \mu S = 0$.Solving for $S$: $S = \frac{h}{\mu}$.Substituting the given values: $S = \frac{1.0}{0.20} = 5 \ m$.

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

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$A$ block of weight $10 \ N$ slides down a curved track $AB$ and then onto a rough horizontal surface. The coefficient of kinetic friction between the block and the rough surface is $0.20$. If the block starts sliding from a height of $1.0 \ m$ above the horizontal surface,calculate the distance $S$ it travels on the rough surface before coming to rest. [$g = 10 \ m \ s^{-2}$]

A
$1$
B
$8$
C
$2$
D
$5$

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Work, Energy, Power and Collision

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Mix Examples-Work, Energy, Power and Collision

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DifficultIIT 1977

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DifficultAIPMT 2012

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$A$ mass of $1 \; kg$ collides elastically with a stationary mass of $5 \; kg$. After the collision,the $1 \; kg$ mass moves in the opposite direction with a velocity of $2 \; ms^{-1}$. Which of the following statements is correct for this system of two masses?

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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 final velocity of the wedge $v_2$ is

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