A small disc of mass m = 1 ,g slides down a s | vedclass

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(C) $1$. First, we calculate the velocity $v$ of the disc at the bottom of the hill using the conservation of mechanical energy:$mgh = \frac{1}{2}mv^2

Vedclass · Accepted Answer

$0.1$

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(C) $1$. First, we calculate the velocity $v$ of the disc at the bottom of the hill using the conservation of mechanical energy:$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}$Given $h = 10 \,cm = 0.1 \,m$ and $g = 10 \,m/s^2$, we have $v = \sqrt{2 	imes 10 	imes 0.1} = \sqrt{2} \,m/s$.$2$. When the disc gets onto the plank, they move together with a common velocity $v'$. By the law of conservation of linear momentum:$mv = (m + M)v' \implies v' = \frac{mv}{m+M}$$3$. The work done by the frictional force $(W_f)$ is equal to the change in kinetic energy of the system:$W_f = K_{final} - K_{initial} = \frac{1}{2}(m+M)v'^2 - \frac{1}{2}mv^2$Substituting $v' = \frac{mv}{m+M}$:$W_f = \frac{1}{2}(m+M) \left(\frac{mv}{m+M}ight)^2 - \frac{1}{2}mv^2 = \frac{1}{2}mv^2 \left(\frac{m}{m+M} - 1ight) = -\frac{1}{2}mv^2 \left(\frac{M}{m+M}ight)$$4$. The magnitude of work done by friction is:$|W_f| = \frac{1}{2}mv^2 \left(\frac{M}{m+M}ight) = mgh \left(\frac{M}{m+M}ight)$Substituting the values $m = 1 \,g = 10^{-3} \,kg$, $M = 100 \,g = 100 	imes 10^{-3} \,kg$, $g = 10 \,m/s^2$, $h = 0.1 \,m$:$|W_f| = (10^{-3}) 	imes 10 	imes 0.1 	imes \left(\frac{100 	imes 10^{-3}}{101 	imes 10^{-3}}ight) = 10^{-3} 	imes \frac{100}{101} \approx 0.99 	imes 10^{-3} \,J \approx 0.001 \,J$.*Correction Note:* Based on the provided options and standard interpretation of such problems, if $h$ were $10 \,m$ instead of $10 \,cm$, the result would be $0.1 \,J$. Given the options, $0.001 \,J$ is the calculated value for $10 \,cm$, but $0.1 \,J$ is the intended answer for $10 \,m$.

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The block starts from rest as shown in the figure. Find the work done by the force of $10 \,N$ and friction in the time $0$ to $4 \,s$. [Take, $g=10 \,m/s^2$]

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