A block of mass $M_{1} = 10 kg$ is placed on a slab of mass $M_{2} = 30 kg$ . The slab lies on a friction less horizontal surface as shown in figure. The coefficient of static friction between the block and slab is $μ_{1} = 0.4$ and that of dynamic friction is $μ_{2} = 0.15$ . The acceleration with which the slab will move is $(g = 10 {ms}^{- 1})$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1.25 {ms}^{- 2}

2 {ms}^{- 2}

0.5 {ms}^{- 2}

5 {ms}^{- 2}

answer is C.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

a_{c m} = \frac{5 ϕ}{4 ϕ} = \frac{5}{4} m / s^{2}

$f = 30 \times \frac{5}{4} = \frac{150}{4} = 37.5$

$f_{l i n} = 0.4 \times 100 = 40 N$

$f < f_{\lim}$ Hence both move together

$∴ a_{s l a b} = \frac{5}{4} = 1.25 m / s^{2}$

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test