A 30 kg slab B rests on a frictionless floor | vedclass

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(D) $1$. Calculate the maximum static frictional force between block $A$ and slab $B$: $f_{s,max} = \mu_s N = \mu_s m_A g = 0.60 	imes 10 	imes 9.8

Vedclass · Accepted Answer

$1.31$

Vedclass · Answer

(D) $1$. Calculate the maximum static frictional force between block $A$ and slab $B$: $f_{s,max} = \mu_s N = \mu_s m_A g = 0.60 	imes 10 	imes 9.8 = 58.8 \ N$. $2$. Since the applied force $(100 \ N)$ is greater than the maximum static friction $(58.8 \ N)$, block $A$ will slide over slab $B$. $3$. Once sliding occurs, the kinetic friction force acts between the surfaces: $f_k = \mu_k N = \mu_k m_A g = 0.40 	imes 10 	imes 9.8 = 39.2 \ N$. $4$. This kinetic friction force $f_k$ is the only horizontal force acting on slab $B$ (since the floor is frictionless). $5$. Using Newton's second law for slab $B$: $F_{net} = M_B a_B \Rightarrow f_k = M_B a_B$. $6$. $39.2 = 30 	imes a_B \Rightarrow a_B = \frac{39.2}{30} \approx 1.31 \ m \ s^{-2}$.

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