(A) According to the work-energy theorem, the work done by the retarding force is equal to the change in kinetic energy of the bus.$W = \Delta K$$F \c

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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 bus.$W = \Delta K$$F \cdot s = \frac{1}{2} m v^2$Since the retarding force $F$ and the initial velocity $v$ are constant, we have:$s = \frac{m v^2}{2 F} \implies s \propto m$Let the initial mass be $m_1 = m$ and the initial stopping distance be $s_1 = x$.The new mass after increasing by $25\%$ is $m_2 = m + 0.25m = 1.25m$.Using the proportionality $s \propto m$:$\frac{s_2}{s_1} = \frac{m_2}{m_1}$$s_2 = s_1 \cdot \frac{1.25m}{m}$$s_2 = 1.25 x$.

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

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$A$ bus moving on a level road with a velocity $v$ can be stopped at a distance of $x$ by the application of a retarding force $F$. The load on the bus is increased by $25\%$ by boarding the passengers. Now, if the bus is moving with the same speed and if the same retarding force is applied, the distance travelled by the bus before it stops is

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$1.25 x$
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$x$
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$5 x$
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$2.5 x$

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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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$A$ bullet of mass $0.04\; kg$ moving with a speed of $90\; m/s$ enters a heavy wooden block and is stopped after a distance of $60\; cm$. What is the average resistive force exerted by the block on the bullet (in $; N$)?

The pulley and spring are massless,and friction is absent everywhere. $A$ $5\, kg$ block is released from rest. Find the speed of the $5\, kg$ block when the $2\, kg$ block leaves contact with the ground. (Take the force constant of the spring $K = 40\, N/m$ and $g = 10\, m/s^2$.)

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