A long plank P of mass 5, kg is placed on a s | vedclass

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(D) $1$. Maximum static friction force between $P$ and $Q$ is $f_{max} = \mu N = 0.5 	imes (2 	imes 10) = 10\, N$.$2$. The applied force on $Q$ is $

Vedclass · Accepted Answer

Both $(B)$ and $(C)$.

Vedclass · Answer

(D) $1$. Maximum static friction force between $P$ and $Q$ is $f_{max} = \mu N = 0.5 	imes (2 	imes 10) = 10\, N$.$2$. The applied force on $Q$ is $F = 15\, N$. Since $F > f_{max}$,the block $Q$ will slide over the plank $P$,and kinetic friction $f_k = 10\, N$ will act between them.$3$. Acceleration of $Q$: $a_Q = (F - f_k) / m_Q = (15 - 10) / 2 = 2.5\, m/s^2$.$4$. Acceleration of $P$: $a_P = f_k / m_P = 10 / 5 = 2.0\, m/s^2$. Thus,statement $(C)$ is correct.$5$. Acceleration of the centre of mass: $a_{cm} = (F_{ext}) / (m_P + m_Q) = 15 / (5 + 2) = 15/7\, m/s^2$. Thus,statement $(B)$ is correct.$6$. The reaction force on $Q$ due to $P$ is the normal force $N = m_Q g = 20\, N$,not $10\, N$. Thus,$(A)$ is incorrect.$7$. Since both $(B)$ and $(C)$ are correct,the correct option is $(D)$.

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