Question

2.
Two blocks with masses M, and M, of
10kg and 20kg respectively are placed as
in fig. m=0.2 between all surfaces, then
tension in string and acceleration of M,
block will be:
30°
A) 250 N, 3 m/s2 B) 200 N, 6 m/s
C) 306 N, 4.7 m/s2 D) 400 N, 6.5 m/s?​

Answer 1

Answer:

Two blocks with masses m1 and m2 of 10kg and 20kg respec are placed as in fig. mew=0.2 bet all surfaces then tension in string and acceleration of m2 block at this moment will be. photo....see ok.

From the m1 block, forces acting on vertical direction, we have, T sin30 = 10g ............(1)

from eqn.(1), we get Tension in the string, T = 196 N

Horizontal componenet of Tension of the string pressing m2 block.

Hence normal reaction force N = Tcos30 = 196×(√3/2) = 170 N

for the m2 block, using the vertical direction forces, Newton's law is written as :

20×g - 2×μ×N = 20×a ....................(2)

by substituting values for μ=0.2, N=170 , we get a = 6.4 m/s2

Answer 2

CORRECT QUESTION :

Two blocks with masses $M_1$ and $M_2$ of 10 kg and 20 kg respectively are placed as in figures. Coefficient of friction μ = 0.2 between all surfaces, then tension in string and acceleration of $M_2$ block will be :

$\huge{\underline{\underline{\boxed{\mathfrak{\red{A}\green{n}\pink{s}\orange{w}\blue{e}{r}}}}}}$

GIVEN :

• $M_1$ = 10 kg.
• $M_2$ = 20 kg.
• μ = 0.2

TO FIND :

• Tension in the string = ?
• Acceleration of $M_2$ block will be = ?

FORMULAS USED :

• T sin 30° = mass × g.
• N = T cos 30°.

SOLUTION :

To find the tension in string,

T sin 30° = mass × g.

T × 0.5 = 10 × 9.8 m/s².

T = 196 N.

•°• Tension in string = 196 N.

Now, the total force of equilibrium,

N = T cos 30°.

N = 196 × √3/2

N = 169.7 N.

•°• Force at equilibrium (N) = 169.7 N.

Now, for block $M_2$,

→ 20 × g - 2 × μ × N = 20 × a.

Now, substituting the values.

we get,

→ 20 × 9.8 - 2 × 0.2 × 169.7 = 20 × a.

→ 20 a = 128.12

→ a = 128.12/20

•°• $\red {\boxed{\textsf a = 6.40 m/s²}}$