A 40 ,kg slab rests on a frictionless floor a | vedclass

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(A) $1$. Calculate the limiting friction between the block $(m_A = 10 \,kg)$ and the slab $(m_B = 40 \,kg)$:$f_{s,max} = \mu_s m_A g = 0.60 	imes 10

Vedclass · Accepted Answer

$0.98$

Vedclass · Answer

(A) $1$. Calculate the limiting friction between the block $(m_A = 10 \,kg)$ and the slab $(m_B = 40 \,kg)$:$f_{s,max} = \mu_s m_A g = 0.60 	imes 10 	imes 9.8 = 58.8 \,N$.$2$. Compare the applied force $(F = 100 \,N)$ with the limiting friction:Since $F > f_{s,max}$ $(100 \,N > 58.8 \,N)$,the block will slip over the slab.$3$. Calculate the kinetic friction force acting between the block and the slab:$f_k = \mu_k m_A g = 0.40 	imes 10 	imes 9.8 = 39.2 \,N$.$4$. This kinetic friction force $f_k$ acts on the slab in the direction of the applied force,causing it to accelerate:$F_{net, slab} = f_k = 39.2 \,N$.$5$. Calculate the acceleration of the slab $(a_B)$:$a_B = \frac{F_{net, slab}}{m_B} = \frac{39.2}{40} = 0.98 \,m/s^2$.

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