Coefficient of kinetic friction between 3 kg and 2 kg block is 0.25 . The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass 10 kg .

Question

Coefficient of kinetic friction between 3 kg and 2 kg block is 0.25 . The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass 10 kg . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-s002avkcgqt5m9vn.jpg" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

100-T2=10a T2-T1-0.25× 20=3a T1-0.25× 20=2a Adding the above three equations and solving for acceleration we get a=(100 - 0.25 × 20 - 0.25 × 20/15)=6 ms- 2

Q. Coefficient of kinetic friction between $3 k g$ and $2 k g$ block is $0.25$ . The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass $10 k g$ .

$100 - T_{2} = 10 a$
$T_{2} - T_{1} - 0.25 \times 20 = 3 a$
$T_{1} - 0.25 \times 20 = 2 a$
Adding the above three equations and solving for acceleration we get
$a = \frac{100 - 0.25 \times 20 - 0.25 \times 20}{15} = 6 m s^{- 2}$

Q. Coefficient of kinetic friction between 3kg and 2kg block is 0.25 . The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass 10kg .

100−T2​=10a T2​−T1​−0.25×20=3a T1​−0.25×20=2a Adding the above three equations and solving for acceleration we get a=15100−0.25×20−0.25×20​=6ms−2

Q. Coefficient of kinetic friction between $3 k g$ and $2 k g$ block is $0.25$ . The horizontal table surface is smooth. Find the acceleration (in S.I unit) of the block of mass $10 k g$ .

$100 - T_{2} = 10 a$
$T_{2} - T_{1} - 0.25 \times 20 = 3 a$
$T_{1} - 0.25 \times 20 = 2 a$
Adding the above three equations and solving for acceleration we get
$a = \frac{100 - 0.25 \times 20 - 0.25 \times 20}{15} = 6 m s^{- 2}$