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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $(a_B)$:$a_B = \frac{F_{net, slab}}{m_B} = \frac{39.2}{40} = 0.98 \,m/s^2$.

Similar Questions

Find the maximum force $F$ (in $N$) for which both the blocks will move with the same acceleration.

$A$ block of mass $m$ is placed on the top of another block of mass $M$ as shown in the figure. The coefficient of friction between them is $\mu$. What is the maximum acceleration with which the block $M$ may move so that $m$ also moves along with it?

The system shown in the figure is released. Find the accelerations $a_1$ (for the $2 \ kg$ block) and $a_2$ (for the $3 \ kg$ block).

Vedclass Test Series

Exam Paper Generator

Online Exam Module